【CodeForces - 520C】DNA Alignment （快速幂，思维）

题干：

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 function h(s, t)is defined as the number of positions in which the respective symbols of s and t are the same. Function h(s, t) can be used to define the function of Vasya distance ρ(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 modulo 109 + 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.

Examples

Input

1
C

Output

Input

2
AG

Output

Input

3
TTT

Output

Note

Please note that if for two distinct strings t1 and t2 values ρ(s, t1) и ρ(s, t2)are maximum among all possible t, 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 strings t of length 1 the value of ρ(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

解题报告：

直接看哪个字母出现的次数最多就行了，注意有可能很多字母出现的次数一样多，所以扫两遍就做完了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
char s[MAX];
int bk[20];
ll mod = 1e9+7;
ll qpow(ll a,ll k) {ll res = 1;while(k) {if(k&1) res = (res*a)%mod;k>>=1;a=(a*a)%mod; }return res;
}
int main()
{ll n;cin>>n;cin>>(s+1);int len = strlen(s+1);for(int i = 1; i<=len; i++) {if(s[i] == 'A') bk[1]++;if(s[i] == 'C') bk[2]++;if(s[i] == 'G') bk[3]++;if(s[i] == 'T') bk[4]++;}int maxx = *max_element(bk+1,bk+5);ll cur = 0;for(int i = 1; i<=4; i++) {if(bk[i] == maxx) cur++;}printf("%lld\n",qpow(cur,n));return 0 ;}

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