#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
const int N_MAX = 1000000000 + 2,INF= 1000000007;
int N;
string s;
vector<pair<int,int> >loop;//圈的大小<->圈的个数

ll gcd(ll a,ll b) {
    if (b == 0)return a;
    return gcd(b, a%b);
}

ll lcm(ll a,ll b) {
    return a / gcd(a, b)*b;
}

void cal_loop() {
    loop.clear();
    int change[26],vis[26];
    memset(vis,0,sizeof(vis));
    for (int i = 0; i < s.size();i++) {
        change[i] = s[i] - ‘a‘;
    }
    map<int, int>m;
    for (int i = 0; i < 26;i++) {
        
        if (vis[i])continue;
        vis[i] = true;
        int num = 1;
        int k = change[i];//k代表不断变化的字母
        while (i != k) {
            vis[k] = true;
            num++;//该圈的元素个数加1
            k = change[k];//!!!!!顺序
        }
        m[num]++;
    }
    for (map<int, int>::iterator it = m.begin(); it != m.end();it++) {
        loop.push_back(*it);
    }
}

ll mod_pow(ll x,ll n) {//快速幂运算
    ll res = 1;
    while (n>0) {
        if (n & 1)res = res*x%INF;
        x = x*x%INF;
        n >>= 1;
    }
    return res;
}

ll R_C(vector<int>&loop,int N) {//容斥原理，求由这些圈中的元素组成的长度为N的字符串的数量
    ll permute = 1 << (loop.size());
    ll ans = 0;
    for (int i = 0; i < permute;i++) {
        int num = 0;
        int sum = 0;
        int sign=-1;
        for (int j = 0; j < loop.size(); j++) {
            if (i&(1 << j)) {
                num++;//num记录利用到的圈的个数
                sum += loop[j];//利用到的字符的总个数
            }
        }    
            if (num % 2 == loop.size() % 2)//!!!!!!!!
                sign = 1;
            ans =(ans+((sign*mod_pow(sum, N))%INF+INF)%INF)%INF;
    }
    return ans;
}

ll solve(int N) {
    cal_loop();
    vector<int>vec;
    ll ans = 0;
    for (int i = 0; i < (1<<loop.size());i++) {
        ll change_time=1;
        vec.clear();
        for (int j = 0; j < loop.size();j++) {
            if (i&(1 << j)) {
                vec.push_back(loop[j].first*loop[j].second);
                change_time = lcm(change_time, loop[j].first);
            }
        }
        if (vec.size() > N)continue;//挑选出来的圈的个数不能超过字符串长度
    ll number = R_C(vec, N);
    ans = (ans + change_time*number) % INF;
    }
    return ans;
}


int main() {
    int T;
    scanf("%d",&T);
    while (T--) {
        scanf("%d",&N);
        cin >> s;
        printf("%lld\n",solve(N));
    }
    return 0;
}