Tinkoff Internship Warmup Round 2018 and Codeforces Round #475 (Div. 2)

A. Splits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's define a split of $\mathrm{nn as\; a\; nonincreasing\; sequence\; of\; positive\; integers,\; the\; sum\; of\; which\; is nn.}$

For example, the following sequences are splits of $\mathrm{88: [4,4][4,4], [3,3,2][3,3,2], [2,2,1,1,1,1][2,2,1,1,1,1], [5,2,1][5,2,1].}$

The following sequences aren't splits of $\mathrm{88: [1,7][1,7], [5,4][5,4], [11,-3][11,-3], [1,1,4,1,1][1,1,4,1,1].}$

The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1,1,1,1,1][1,1,1,1,1] is 55,\; the\; weight\; of\; the\; split [5,5,3,3,3][5,5,3,3,3] is 22 and\; the\; weight\; of\; the\; split [9][9] equals 11.$

For a given $\mathrm{nn,\; find\; out\; the\; number\; of\; different\; weights\; of\; its\; splits.}$

Input

The first line contains one integer $\mathrm{nn (1\le n\le 1091\le n\le 109).}$

Output

Output one integer — the answer to the problem.

Examples

input

Copy

7

output

Copy

4

input

Copy

8

output

Copy

5

input

Copy

9

output

Copy

5

Note

In the first sample, there are following possible weights of splits of $\mathrm{77:}$

Weight 1: [$\text{77]}$

Weight 2: [$\text{33, 33, 1]}$

Weight 3: [$\text{22, 22, 22, 1]}$

Weight 7: [$\text{11, 11, 11, 11, 11, 11, 11]}$

#include <iostream>
using namespace std;

int main(){
    int n;
    cin>>n;
    n=n>>1;
    cout<<n+1<<endl;
    return 0;
}

B. Messages

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n incoming messages for Vasya. The i-th message is going to be received after ti minutes. Each message has a cost, which equals to A initially. After being received, the cost of a message decreases by B each minute (it can become negative). Vasya can read any message after receiving it at any moment of time. After reading the message, Vasya's bank account receives the current cost of this message. Initially, Vasya's bank account is at 0.

Also, each minute Vasya's bank account receives C·k, where k is the amount of received but unread messages.

Vasya's messages are very important to him, and because of that he wants to have all messages read after T minutes.

Determine the maximum amount of money Vasya's bank account can hold after T minutes.

Input

The first line contains five integers n, A, B, C and T (1 ≤ n, A, B, C, T ≤ 1000).

The second string contains n integers ti (1 ≤ ti ≤ T).

Output

Output one integer  — the answer to the problem.

Examples

input

Copy

4 5 5 3 5
1 5 5 4

output

Copy

20

input

Copy

5 3 1 1 3
2 2 2 1 1

output

Copy

15

input

Copy

5 5 3 4 5
1 2 3 4 5

output

Copy

35

Note

In the first sample the messages must be read immediately after receiving, Vasya receives A points for each message, n·A = 20 in total.

In the second sample the messages can be read at any integer moment.

In the third sample messages must be read at the moment T. This way Vasya has 1, 2, 3, 4 and 0 unread messages at the corresponding minutes, he gets 40 points for them. When reading messages, he receives (5 - 4·3) + (5 - 3·3) + (5 - 2·3) + (5 - 1·3) + 5 =  - 5 points. This is 35 in total.

#include <iostream>
using namespace std;

int n,A,B,C,T;
int a[1005];
int main(){
    cin>>n>>A>>B>>C>>T;
    for(int i=0;i<n;i++){
        int x;
        cin>>x;
        a[x]++;
    }
    for(int i=1;i<=T;i++)
        a[i]+=a[i-1];
    if(B>=C){
        cout<<A*n<<endl;
    }else{
        int sum = 0;
        for(int i=T-1;i>=1;i--){
            sum+=a[i];
        }
        sum = sum*(C-B);
        cout<<A*n+sum<<endl;
    }

    return 0;
}

 

C. Alternating Sum

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two integers $\mathrm{aa and bb.\; Moreover,\; you\; are\; given\; a\; sequence s0,s1,\dots ,sns0,s1,\dots ,sn.\; All\; values\; in ss are\; integers 11 or -1-1.\; It\text{'}s\; known\; that\; sequence\; is kk-periodic\; and kk divides n+1n+1.\; In\; other\; words,\; for\; each k\le i\le nk\le i\le n it\text{'}s\; satisfied\; that si=si-ksi=si-k.}$

Find out the non-negative remainder of division of $\stackrel{}{}$

Note that the modulo is unusual!

Input

The first line contains four integers $\mathrm{n,a,bn,a,b and kk (1\le n\le 109,1\le a,b\le 109,1\le k\le 105)(1\le n\le 109,1\le a,b\le 109,1\le k\le 105).}$

The second line contains a sequence of length $\mathrm{kk consisting\; of\; characters\; \text{'}+\text{'}\; and\; \text{'}-\text{'}.}$

If the $\mathrm{ii-th\; character\; (0-indexed)\; is\; \text{'}+\text{'},\; then si=1si=1,\; otherwise si=-1si=-1.}$

Note that only the first $\mathrm{kk members\; of\; the\; sequence\; are\; given,\; the\; rest\; can\; be\; obtained\; using\; the\; periodicity\; property.}$

Output

Output a single integer — value of given expression modulo ${\mathrm{109+9109+9.}}^{}$

Examples

input

Copy

2 2 3 3
+-+

output

Copy

7

input

Copy

4 1 5 1
-

output

Copy

999999228

Note

In the first example:

$(n\sum i=0sian-ibi)(\sum i=0nsian-ibi) = 2230-2131+20322230-2131+2032 =\; 7$

In the second example:

$(n\sum i=0sian-ibi)=-1450-1351-1252-1153-1054=-781\equiv 999999228(mod109+9)(\sum i=0nsian-ibi)=-1450-1351-1252-1153-1054=-781\equiv 999999228(mod109+9).$

快速幂，加逆元，再加等比数列求和。

#include <iostream>
#include <bits/stdc++.h>
#define ll long long int
#define Mod 1000000009
using namespace std;

ll quick(ll a,ll b){
    ll res = 1;
    while(b>0){
        if(b&1){
            res=res*a%Mod;
        }
        b>>=1;
        a=a*a%Mod;
    }
    return res%Mod;
}

ll n,a,b,k;
string s;

int main(){
    cin>>n>>a>>b>>k;
    cin>>s;
    ll ans = 0;
    for(int i=0;i<k;i++){
        ll pink = s[i]=='+'?1:-1;
        ans=(ans+pink*quick(a,n-i)*quick(b,i)%Mod+Mod)%Mod;
    }
    ll cnt = b*quick(a,Mod-2)%Mod;//求a的逆元
    cnt = quick(cnt,k);
    ll num = (n+1)/k;
    ll sum;
    if(cnt == 1){
        sum = num*ans%Mod;
    }else{
        sum = (ans*(quick(cnt,num)-1)%Mod*quick(cnt-1,Mod-2)+Mod)%Mod;
    }
    cout<<sum<<endl;
    return 0;
}