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 <stdio.h>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
using ll = long long;
constexpr ll mod = 1e9 + 7;

ll fact[2100010], ifact[2100010];
ll mpow(ll n, ll e) {
	ll r = 1;
	while (e) {
		if (e & 1)
			r = r * n % mod;
		n = n * n % mod;
		e /= 2;
	}
	return r;
}
void init() {
	fact[0] = 1;
	for (int i = 1; i <= 2100000; i++)
		fact[i] = fact[i - 1] * i % mod;
	ifact[2100000] = mpow(fact[2100000], mod - 2);
	for (int i = 2099999; 0 <= i; i--)
		ifact[i] = ifact[i + 1] * (i + 1) % mod;
}
ll comb(ll n, ll r) {
	return fact[n] * ifact[n - r] % mod * ifact[r] % mod;
}
ll k, n;
char s[1100000];
int main() {
	scanf("%lld", &k);
	scanf("%s", s);
	for (; s[n]; n++);
	init();
	ll sum = mpow(26, k + n);
	for (int i = 0; i < n; i++) {
		sum -= mpow(25, k + n - i) * comb(k + n, i) % mod;
		sum = (sum + mod) % mod;
	}
	printf("%lld\n", sum);
}