#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define ff first
#define ss second
#define vll vector<ll>
#define pll pair<ll,ll>
#define vpll vector<pll>
#define vvll vector<vll>
#define all(v) v.begin(),v.end()
#define itr(r,v) for(auto r=v.begin();r!=v.end();r++)
#define W while
#define FOR(i,n) for(ll i=0;i<n;i++)
#define ffor(i,a,b) for(ll i=a;i<=b;i++)
#define rfor(i,a,b) for(ll i=a;i>=b;i--)
#define space cout<<"\n\n";
#define endl "\n"
#define pqmx priority_queue<ll>
#define pqmn priority_queue<ll,vll,greater<ll>>
#define fps(x,y) fixed<<setprecision(y)<<x
#define merg(a,b,c) set_union(a.begin(),a.end(),b.begin(),b.end(),inserter(c,c.begin()))
#define set_ar(arr,v) memset(arr,v,sizeof(arr))
#define ctime auto start = high_resolution_clock::now()
#define etime auto stop = high_resolution_clock::now()
#define ptime auto z1z = duration_cast<microseconds>(stop-start); cout<<"Time elapsed : "<<z1z.count()<<" microseconds\n"
#define go_t ll testcases; cin>>testcases; while(testcases--)
#define dbg1(x)             cout <<#x<<": "<<x<<'\n';
#define dbg2(x, y)          cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<'\n';
#define dbg3(x, y, z)       cout <<#x<<": "<<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<'\n';
#define dbg4(a, b, c, d)    cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<'\n';
#define dbg5(a, b, c, d, e) cout <<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<<": "<<e<<'\n';
using namespace std;
using namespace std::chrono;

const ll MOD = 1e9 + 7;
const ll N = 2e6 + 6;
const ll maxN = 1e5 + 5;
const ll MAX_SIZE = 1e8 + 8;

ll power(ll x, ll y, ll M = MOD) // default argument
{ll v = 1; x = x % M; while (y > 0) {if (y & 1)v = (v * x) % M; y = y >> 1; x = (x * x) % M;} return v;}

ll modinv(ll a) { return power(a, MOD - 2, MOD); }

ll fact[N], invfact[N];

ll nCr(ll x, ll y)
{
	if (y > x)
		return 0;
	ll num = fact[x];
	num *= invfact[y];
	num %= MOD;
	num *= invfact[x - y];
	num %= MOD;
	return num;
}

void precompute()
{
	fact[0] = fact[1] = 1;
	for (ll i = 2; i < N; i++)
	{
		fact[i] = fact[i - 1] * i;
		fact[i] %= MOD;
	}
	/*
	n! = n * (n-1) !
	==> 1/(n-1!) = n/(n!)
	==> (1/(n-1!)) % M = (n/(n!)) % M
	==> (1/(n-1!)) % M = ((n % M) * ((1/(n!)) % M)) % M
	*/
	invfact[N - 1] = modinv(fact[N - 1]);
	for (ll i = N - 2; i >= 0; i--)
	{
		invfact[i] = invfact[i + 1] * (i + 1);
		invfact[i] %= MOD;
	}

}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("input1.txt", "r", stdin);
	freopen("output1.txt", "w", stdout);
#endif
	// WHILE(!(SUCCEED == TRY()));
	// BEFORE WRITING MAKE SURE YOU HAVE ANALYSED THE CONSTRAINTS
	// AND THINK THROUGH THE LOGIC ONCE MORE
	FIO
	precompute();
	ll k; cin >> k;
	string s; cin >> s;
	ll leng = s.length() + k;
	ll tot = power(26, leng, MOD);
	// tot contains all the possible number of strings of leng that are possible
	ll all_not_poss = 0;
	for (ll i = 0; i < s.length(); i++) {
		all_not_poss += (nCr(leng, i) * power(25, leng - i, MOD)) % MOD ;
	}
	ll ans = ((tot - all_not_poss) % MOD + MOD ) % MOD;
	cout << ans;
	return 0;
}
/*
https://codeforces.com/blog/entry/79153

Let's consider the end result of this process and what it can look like. It will be a string (let's call it T) of length N=K+|S| that contains S

as a subsequence.

We'll use Z=26

for readability.

In order to count these, let's first note that there are ZN
total strings of length N, and we want to remove all the strings that don't contain S as a subsequence (this is a bit easier than counting the strings that do contain S, because a string could contain S

multiple times and be overcounted).

Let's consider the length of the maximum prefix of S
that's a subsequence of our string T. Going from left to right within T, we can check when this increases -- it increases from X to X+1 when our current character is SX+1. In order for it to not increase, we need our current character to not be SX+1

.

So let's say this length is exactly M
. Then it must have increased from 0 to M (one at a time, of course), so we pick M indices in T to be the indices where it increases, and those have only one choice of character. The remaining characters have Z−1 choices. So for length M, we have (NM)(Z−1)N−M

possibilities.

In order to not contain S
as a subsequence, M can be anything from 0 to |S|−1

. Now, recall these are the strings we don't want, so we subtract them from the total.

Therefore, our final answer is (Z^N) - summation(m)(from 0 to |S|-1)((N C m) * ((Z-1)^(N-m)))


Runtime: O(K+|S|)
.*/


Measured memory_bytes: 35516