Codeforces Round #502 (in memory of Leopoldo Taravilse, Div. 1 + Div. 2) ABC

A. The Rank

John Smith knows that his son, Thomas Smith, is among the best students in his class and even in his school. After the students of the school took the exams in English, German, Math, and History, a table of results was formed.

There are $\mathrm{nn students,\; each\; of\; them\; has\; a unique id\; (from 11 to nn).\; Thomas\text{'}s\; id\; is 11.\; Every\; student\; has\; four\; scores\; correspond\; to\; his\; or\; her\; English,\; German,\; Math,\; and\; History\; scores.\; The\; students\; are\; given\; in\; order\; of\; increasing\; of\; their\; ids.}$

In the table, the students will be sorted by decreasing the sum of their scores. So, a student with the largest sum will get the first place. If two or more students have the same sum, these students will be sorted by increasing their ids.

Please help John find out the rank of his son.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 10001\le n\le 1000) —\; the\; number\; of\; students.}$

Each of the next $\mathrm{nn lines\; contains\; four\; integers aiai, bibi, cici,\; and didi (0\le ai,bi,ci,di\le 1000\le ai,bi,ci,di\le 100) —\; the\; grades\; of\; the ii-th\; student\; on\; English,\; German,\; Math,\; and\; History.\; The\; id\; of\; the ii-th\; student\; is\; equal\; to ii.}$

Output

Print the rank of Thomas Smith. Thomas's id is $11.$

Examples

input

Copy

5
100 98 100 100
100 100 100 100
100 100 99 99
90 99 90 100
100 98 60 99

output

2

input

6
100 80 90 99
60 60 60 60
90 60 100 60
60 100 60 80
100 100 0 100
0 0 0 0

output

1

Note

In the first sample, the students got total scores: $\mathrm{398398, 400400, 398398, 379379,\; and 357357.\; Among\; the 55 students,\; Thomas\; and\; the\; third\; student\; have\; the\; second\; highest\; score,\; but\; Thomas\; has\; a\; smaller\; id,\; so\; his\; rank\; is 22.}$

In the second sample, the students got total scores: $\mathrm{369369, 240240, 310310, 300300, 300300,\; and 00.\; Among\; the 66 students,\; Thomas\; got\; the\; highest\; score,\; so\; his\; rank\; is 11.}$

求第一个人的排名。

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
struct Nod{
    int id,s;
}e[N];
int a, n, b, c, d;
bool cmp(const Nod &a, const Nod &b) {
    if(a.s != b.s) return a.s > b.s;
    else return a.id < b.id;
}
int main() {
    cin >> n;
    for(int i = 1; i <= n; i ++) {
        cin >> a >> b >> c >> d;
        e[i].id = i;
        e[i].s = a+b+c+d;
    }
    sort(e+1,e+1+n,cmp);
    for(int i = 1; i <= n; i ++) 
        if(e[i].id == 1) return 0*printf("%d
",i);
    return 0;
}

B. The Bits

Rudolf is on his way to the castle. Before getting into the castle, the security staff asked him a question:

Given two binary numbers $\mathrm{aa and bb of\; length nn.\; How\; many\; different\; ways\; of\; swapping\; two\; digits\; in aa (only\; in aa,\; not bb)\; so\; that\; bitwise\; OR\; of\; these\; two\; numbers\; will\; be\; changed?\; In\; other\; words,\; let cc be\; the\; bitwise\; OR\; of aa and bb,\; you\; need\; to\; find\; the\; number\; of\; ways\; of\; swapping\; two\; bits\; in aa so\; that\; bitwise\; OR\; will\; not\; be\; equal\; to cc.}$

Note that binary numbers can contain leading zeros so that length of each number is exactly $\mathrm{nn.}$

Bitwise OR is a binary operation. A result is a binary number which contains a one in each digit if there is a one in at least one of the two numbers. For example, ${\mathrm{010102010102 OR 100112100112 = 110112110112.}}^{}$

Well, to your surprise, you are not Rudolf, and you don't need to help him$\dots \dots You\; are\; the\; security\; staff!\; Please\; find\; the\; number\; of\; ways\; of\; swapping\; two\; bits\; in aa so\; that\; bitwise\; OR\; will\; be\; changed.$

Input

The first line contains one integer $\mathrm{nn (2\le n\le 1052\le n\le 105) —\; the\; number\; of\; bits\; in\; each\; number.}$

The second line contains a binary number $\mathrm{aa of\; length nn.}$

The third line contains a binary number $\mathrm{bb of\; length nn.}$

Output

Print the number of ways to swap two bits in $\mathrm{aa so\; that\; bitwise\; OR\; will\; be\; changed.}$

Examples

input

Copy

5
01011
11001

output

Copy

4

input

Copy

6
011000
010011

output

Copy

6

Note

In the first sample, you can swap bits that have indexes $(1,4)(1,4), (2,3)(2,3), (3,4)(3,4),\; and (3,5)(3,5).$

In the second example, you can swap bits that have indexes $(1,2)(1,2), (1,3)(1,3), (2,4)(2,4), (3,4)(3,4), (3,5)(3,5),\; and (3,6)(3,6).$

只能交换a串，那就找b中为0的，因为为1不管怎么交换a串都不会改变原来的值。然后在记录a串中的0、1的数量就行了。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5+10;
char s1[N], s2[N];

int main() {
    ll x = 0, y = 0, x1 = 0, y1 = 0, n;
    cin >> n;
    scanf("%s%s",s1,s2);
    for(int i = 0; i < n; i ++) {
        if(s2[i] == '0' && s1[i] == '0') x++;
        else if(s2[i] == '0' && s1[i] == '1') y++;
        if(s1[i] == '0') x1++;
        else y1++;
    }
    printf("%lld
",x*y1+y*x1-x*y);
    return 0;
}

C. The Phone Number

Mrs. Smith is trying to contact her husband, John Smith, but she forgot the secret phone number!

The only thing Mrs. Smith remembered was that any permutation of $\mathrm{nn can\; be\; a\; secret\; phone\; number.\; Only\; those\; permutations\; that\; minimize\; secret\; value\; might\; be\; the\; phone\; of\; her\; husband.}$

The sequence of $\mathrm{nn integers\; is\; called\; a\; permutation\; if\; it\; contains\; all\; integers\; from 11 to nn exactly\; once.}$

The secret value of a phone number is defined as the sum of the length of the longest increasing subsequence (LIS) and length of the longest decreasing subsequence (LDS).

A subsequence ${\mathrm{ai1,ai2,\dots ,aikai1,ai2,\dots ,aik where 1\le i1<i2<\dots <ik\le n1\le i1<i2<\dots <ik\le n is\; called\; increasing\; if ai1<ai2<ai3<\dots <aikai1<ai2<ai3<\dots <aik.\; If ai1>ai2>ai3>\dots >aikai1>ai2>ai3>\dots >aik,\; a\; subsequence\; is\; called\; decreasing.\; An\; increasing/decreasing\; subsequence\; is\; called\; longest\; if\; it\; has\; maximum\; length\; among\; all\; increasing/decreasing\; subsequences.}}^{}$

For example, if there is a permutation $[6,4,1,7,2,3,5][6,4,1,7,2,3,5], LIS of\; this\; permutation\; will\; be [1,2,3,5][1,2,3,5],\; so\; the\; length\; of LIS is\; equal\; to 44. LDScan\; be [6,4,1][6,4,1], [6,4,2][6,4,2],\; or [6,4,3][6,4,3],\; so\; the\; length\; of LDS is 33.$

Note, the lengths of LIS and LDS can be different.

So please help Mrs. Smith to find a permutation that gives a minimum sum of lengths of LIS and LDS.

Input

The only line contains one integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; length\; of\; permutation\; that\; you\; need\; to\; build.}$

Output

Print a permutation that gives a minimum sum of lengths of LIS and LDS.

If there are multiple answers, print any.

Examples

input

Copy

4

output

Copy

3 4 1 2

input

Copy

2

output

Copy

2 1

Note

In the first sample, you can build a permutation $[3,4,1,2][3,4,1,2]. LIS is [3,4][3,4] (or [1,2][1,2]),\; so\; the\; length\; of LIS is\; equal\; to 22. LDS can\; be\; ony\; of [3,1][3,1], [4,2][4,2], [3,2][3,2],\; or [4,1][4,1].\; The\; length\; of LDS is\; also\; equal\; to 22.\; The\; sum\; is\; equal\; to 44.\; Note\; that [3,4,1,2][3,4,1,2] is not the\; only\; permutation\; that\; is\; valid.$

In the second sample, you can build a permutation $[2,1][2,1]. LIS is [1][1] (or [2][2]),\; so\; the\; length\; of LIS is\; equal\; to 11. LDS is [2,1][2,1],\; so\; the\; length\; of LDS is\; equal\; to 22.\; The\; sum\; is\; equal\; to 33.\; Note\; that\; permutation [1,2][1,2] is\; also\; valid.$

求LIS + LDS. 最小，然后就是求(i+n/i（向上取整）)最小的值。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;

int main() {
    int n, x, y;
    x = y = INF;
    cin >> n;
    for(int i = 1; i <= sqrt(n); i ++) {
        if(i+(n-1)/i+1 < x + y) {
            x = i;
            y = (n-1)/i+1;
        }
    }
    for(int i = x; i >= 1; i --) {
        for(int j = 0; j < y; j ++) {
            int ans = (i-1)*y+1+j;
            if(1 <= ans && ans <= n)printf("%d ",ans);
        }
    }
    printf("
");
    return 0;
}