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

1

Input

2
AG

Output

4

Input

3
TTT

Output

1

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
題目連結
參考題解1
參考題解2
給定長度為n的一個字串s。
構造長度也為n的字串t。使得p(s,t)值最大，問有多少個不同的t，emmm，這個題目看了題解也是沒看懂為什麼這麼做。大體意思好像是，找到裡面出現次數最多的，那幾個字母。因為，你每確定一個字母，那麼排列組合的時候，就會出現在n個位置上，這樣就提供了n的價值，找到出現次數最多的，那麼提供的價值就最大。找出s中出現最多的次數。如果這個次數對應的字母k個，答案就是kn(每個位置可以有k種選擇，n個位置)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAX_N = 100007;
const long long mod = 1e9 + 7;
char str[MAX_N];
int a[5], n;

long long Pow(long long x, long long n)
{
    long long res = 1;
    while (n > 0)
    {
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

int main()
{
    scanf("%d%s", &n, str);
    for (int i = 0; str[i]; ++i)
    {
        if (str[i] == 'A') ++a[0];
        if (str[i] == 'C') ++a[1];
        if (str[i] == 'G') ++a[2];
        if (str[i] == 'T') ++a[3];
    }
    int up = 0;
    for (int i = 0; i < 4; ++i) up = max(up, a[i]);
    int cnt = 0;
    for (int i = 0; i < 4; ++i) if (a[i] == up) ++cnt;
    long long ans = Pow(cnt, n);
    printf("%lld\n", ans);
    return 0;
}