#include <bits/stdc++.h>

using namespace std;

const int N = 2005, P = 998244353;

int nxt[N], ch[N][26], mx[N], f[N][N];

signed main() {
    int m; scanf("%d", &m);
    string S; cin >> S;
    int n = S.size(); S = ' ' + S;
    for (int i = 2, j = 0; i <= n; ++i) {
        while (j && S[i] != S[j + 1]) j = nxt[j];
        if (S[i] == S[j + 1]) ++j;
        nxt[i] = j;
    }
    for (int i = 0; i <= n; ++i) {
        for (int j = 0; j < 26; ++j) {
            if (i != n && S[i + 1] - 'a' == j) ch[i][j] = i + 1;
            else ch[i][j] = ch[nxt[i]][j];
            if (ch[i][j]) mx[i] = j;
        }
    }
    f[0][0] = 1;
    for (int i = 0; i <= m; ++i) {
        for (int j = 0; j <= n; ++j) {
            for (int k = mx[j]; k < 26; ++k) {
                ((f[i + 1][ch[j][k]] += f[i][j]) >= P) && (f[i + 1][ch[j][k]] -= P);
            }
        }
    }
    int res = 0;
    for (int i = 0; i <= n; ++i) {
        int now = i;
        for (int j = 1; j <= m; ++j) {
            res = (res + 1ll * (25 - mx[now]) * f[m - j][i]) % P;
            now = ch[now][mx[now]];
            if (!now) break;
        }
        res += (i != 0 && now == i);
    }
    int mul = 1;
    for (int i = 1; i <= m; ++i) mul = 26ll * mul % P;
    printf("%d\n", (mul - res + P) % P);
    return 0;
}