#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
int p;
const int maxn = 1e2;
char str[maxn];
int quick_pow(int a, int b, int MOD) {  //求解 a^b%MOD的值
    int base = a % MOD;
    int ans = 1; //相乘，所以这里是1
    while (b) {
        if (b & 1) {
            ans = (ans * base) % MOD; //如果这里是很大的数据，就要用quick_mul
        }
        base = (base * base) % MOD;    //notice。注意这里,每次的base是自己base倍
        b >>= 1;
    }
    return ans;
}

class GaussMatrix {  //复杂度O(n3)
public:
    int a[maxn][maxn];
    int equ, val; //方程（行）个数，和变量（列）个数，其中第val个是b值，不能取
    void init() {
        for (int i = 1; i <= equ; ++i) {
            for (int j = 1; j <= val; ++j) {
                a[i][j] = 0.0;
            }
        }
    }
    void swapRow(int rowOne, int rowTwo) {
        for (int i = 1; i <= val; ++i) {
            swap(a[rowOne][i], a[rowTwo][i]);
        }
    }
    void swapCol(int colOne, int colTwo) {
        for (int i = 1; i <= equ; ++i) {
            swap(a[i][colOne], a[i][colTwo]);
        }
    }
    bool same(int x, int y) {
        return x == y;
    }
    int guass() {
        int k, col; // col，当前要处理的列， k当前处理的行
        for (k = 1, col = 1; k <= equ && col < val; ++k, ++col) { //col不能取到第val个
            int maxRow = k; //选出列最大值所在的行，这样使得误差最小。（没懂）
            for (int i = k + 1; i <= equ; ++i) {
                if (abs(a[i][col]) > abs(a[maxRow][col])) {
                    maxRow = i;
                }
            }
            if (same(a[maxRow][col], 0)) { //如果在第k行以后，整一列都是0
                --k; //则这个变量就是一个自由变量。
                continue;
            }
            if (maxRow != k) swapRow(k, maxRow); // k是当前的最大行了
            for (int i = col + 1; i <= val; ++i) { //整一列约去系数
//                a[k][i] /= a[k][col];
                a[k][i] = (a[k][i] * quick_pow(a[k][col], p - 2, p)) % p;
            }
            a[k][col] = 1; //第一个就要变成1了，然后它下面和上面的变成0
            for (int i = 1; i <= equ; ++i) {
                if (i == k) continue; //当前这行，不操作
                for (int j = col + 1; j <= val; ++j) { //要使a[i][col] = 0,则需要a[i][col]倍
//                    a[i][j] -= a[i][col] * a[k][j]; //这一行减去相应的倍数
                    a[i][j] = (a[i][j] - (a[i][col] * a[k][j]) % p + p) % p;
                }
                a[i][col] = 0;
            }
//            debug();
        }
        for (int res = k; res <= equ; ++res) {
            if (!same(a[res][val], 0)) return -1; //方程无解
        }
        return val - k; //自由变量个数
    }
    void debug() {
        for (int i = 1; i <= equ; ++i) {
            for (int j = 1; j <= val; ++j) {
                printf("%d ", a[i][j]);
            }
            printf("\n");
        }
        printf("*******************************************\n\n");
    }
} arr;
void init() {
    arr.init();
    int lenstr = strlen(str + 1);
    arr.equ = lenstr, arr.val = lenstr + 1;
    int now, to = 1;
    for (int i = 1; i <= lenstr; ++i) {
        now = 1;
        for (int j = 1; j <= lenstr; ++j) {
            arr.a[i][j] = now;
            now = now * to % p;
        }
        to++;
        if (str[i] == ‘*‘) arr.a[i][lenstr + 1] = 0;
        else arr.a[i][lenstr + 1] = str[i] - ‘a‘ + 1;
    }
//    arr.debug();
}
void work() {
    cin >> p >> str + 1;
    init();
    int res = arr.guass();
//    assert(res == 0);
    int lenstr = strlen(str + 1);
    for (int i = 1; i <= lenstr; ++i) {
        cout << arr.a[i][lenstr + 1] << " ";
    }
    cout << endl;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    cin >> t;
    while (t--) work();
    return 0;
}