How many strings can be obtained by applying the following operation on a string S exactly K times: "choose one lowercase English letter and insert it somewhere"?

The answer can be enormous, so print it modulo (10^9+7).

Constraints

* K is an integer between 1 and 10^6 (inclusive).
* S is a string of length between 1 and 10^6 (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:


K
S


Output

Print the number of strings satisfying the condition, modulo (10^9+7).

Examples

Input

5
oof


Output

575111451


Input

37564
whydidyoudesertme


Output

318008117
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;
const int N = 2e6 + 10;
char s[N];
int n, k;
ll fac[N], inv[N];
ll power(ll a, int x) {
	ll ans = 1;
	while (x) {
		if (x & 1) ans = (ans * a) % mod;
		a = (a * a) % mod;
		x >>= 1;
	}
	return ans;
}
void init() {
	fac[0] = 1;
	for (int i = 1; i < N; i++) {
		fac[i] = fac[i - 1] * i %mod;
	}
	inv[N - 1] = power(fac[N - 1], mod - 2);
	for (int i = N - 2; i >= 0; i--) {
		inv[i] = inv[i + 1] * (i + 1) % mod;
	}
}
ll C(int n, int k) {
	return fac[n] * inv[k] % mod*inv[n - k] % mod;
}
int main() {
	scanf("%d", &k);
	scanf("%s", s);
	n = strlen(s);
	init();
	ll ans = 0;
	for (int i = 0; i <= k; i++) {
		ans = (ans + power(26, i)*C(n + k - i - 1, n - 1) % mod*power(25, k - i) % mod) % mod;
	}
	printf("%lld\n", ans);
	return 0;
}