CodeForces

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

题意：给定N，输出一个N的排列，满足其LIS+LDS最小。

思路：当时猜到了是根号倒序排序，但是写WA了。 应该是有什么证明的，不过不难想。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
int a[maxn];
int main()
{
    int N,tN,i,j;
    scanf("%d",&N); tN=sqrt(N); int p=N;
    for(i=1;i<=N;i+=tN){
        for(j=min(N,i+tN-1);j>=i;j--) a[j]=p--;
    }
    for(i=1;i<=N;i++) printf("%d ",a[i]);
    return 0;
}