Codeforces 520C. DNA Alignment 找规律 （简单快速幂5）

F - DNA Alignment

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Description

Vasya became interested in bioinformatics. He's going to write an article about similar cyclic DNA sequences, so he invented a new method for determining

 the similarity of cyclic sequences.

Let's assume that strings s and t have the same length n, then the functionh(s, t) is defined as the number of positions in which the respective symbols ofs 

and t arethe same. Functionh(s, t) can be used to define the function of Vasya distanceρ(s, t):

where is obtained from string s, by applying left circular shift i times. For example,

ρ("AGC", "CGT") =  h("AGC", "CGT") + h("AGC", "GTC") + h("AGC", "TCG") +  h("GCA", "CGT") + h("GCA", "GTC") + h("GCA", "TCG") +  h("CAG", "CGT") + h("CAG", "GTC") + h("CAG", "TCG") =  1 + 1 + 0 + 0 + 1 + 1 + 1 + 0 + 1 = 6

Vasya found a string s of length n on the Internet. Now he wants to count how many strings t there are such that the Vasya distance from the string s attains

 maximum possible value. Formally speaking, t must satisfy the equation:.

Vasya could not try all possible strings to find an answer, so he needs your help. As the answer may be very large, count the number of such strings modulo10⁹ + 7.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 10⁵).

The second line of the input contains a single string of length n, consisting of characters "ACGT".

Output

Print a single number — the answer modulo 10⁹ + 7.

Sample Input

Input

     

      1
     
     

      C
     

Output

1

Input

     

      2
     
     

      AG
     

Output

4

Input

     

      3
     
     

      TTT
     

Output

1

Sample Output

Hint

Please note that if for two distinct strings t₁ andt₂ valuesρ(s, t₁) иρ(s, t₂) are maximum among all possiblet, then both strings must be taken into 

account in the answer even if one of them can be obtained by a circular shift of another one.

In the first sample, there is ρ("C", "C") = 1, for the remaining stringst of length 1 the value ofρ(s, t) is 0.

In the second sample, ρ("AG", "AG") = ρ("AG", "GA") = ρ("AG", "AA") = ρ("AG", "GG") = 4.

In the third sample, ρ("TTT", "TTT") = 27

分析：这道题读了好久才弄明白是什么意思，实际上就是求出现最多的次数的字符，若出现次数最多的字母不唯一，那么相当于B串每一位我们都有不唯一的选择，

设我们有x种选择，那么最后构造出的B串种数就是ans^n,因此我们就可以得到最终答案：ans^n 取摸。

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
using namespace std;
#define ll long long
const ll M=1000000007;
int n;
string input;
map<char,ll> a;
const string s="ATCG";
ll maxn=-1,cnt,ans=1;
int main()
{
    cin>>n>>input;
    for(int i=0;i<n;i++)
        a[input[i]]++;
    for(int i=0;i<4;i++)//找出出现次数最多的字符
        if(a[s[i]]==maxn)
            cnt++;
        else if(a[s[i]]>maxn)
        {
            maxn=a[s[i]];
            cnt=1;
        }
    while(n)
    {
    if(n&1)ans*=cnt;
    ans%=M;
    n>>=1;
    cnt=cnt*cnt%M;}

    cout<<ans<<endl;
    return 0;
}