Problem:
Score: 600 points

Problem Statement

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).

Sample Input 1

5
oof

Sample Output 1

575111451

For example, we can obtain proofend, moonwolf, and onionpuf, while we cannot obtain oofsix, oofelevennn, voxafolt, or fooooooo.

Sample Input 2

37564
whydidyoudesertme

Sample Output 2

318008117

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define ld long double
#define pb push_back
#define mp make_pair

const int MAXN=205;
const ll inf = 1e12;
const ll mod=1000000007;

typedef pair<ll,ll> pi;

ll gc(ll a,ll b)
{
    if(b==0) return a;
    return gc(b,a%b);
}
ll power(ll x, ll y, ll p)
{
    ll res = 1;     // Initialize result

    x = x % p; // Update x if it is more than or
                // equal to p

    if (x == 0) return 0; // In case x is divisible by p;

    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;

        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}

vector<bool>prime(1000005,true);
vector<ll>v;
void sieve(ll n)
{
    prime[1]=false;
   for (int p=2; p*p<=n; p++)
    {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true)
        {
            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i=p*p; i<=n; i += p)
                prime[i] = false;
        }
    }

    for(int i=2;i<=n;i++)
    {
        if(prime[i]==true)
            v.push_back(i);
    }
}



inline ll add(ll a,ll b) {return ((a%mod)+(b%mod))%mod;}
inline ll sub(ll a,ll b) {return ((a%mod)-(b%mod)+mod)%mod;}
inline ll mul(ll a,ll b) {return ((a%mod)*(b%mod))%mod;}


ll recur(ll idx,ll sum, vector<pair<ll,ll> >& v, vector<ll>& b, vector<ll>& pre, vector<ll>& dp)
{
    if(idx==b.size()) return 0;
    if(dp[idx]!=-1) return dp[idx];
    ll temp=INT_MAX;
    temp=min(temp,recur(idx+1,sum,v,b,pre,dp) + b[idx]);
    for(int i=0;i<v.size();i++)
    {
        ll x=v[i].first;
        ll y=v[i].second;
        if(idx+x-1>=b.size()) continue;
        ll calc= pre[idx+x-1]-pre[idx+y-1];
        temp=min(temp, recur(idx+x, sum,v,b,pre,dp)+ calc);
    }
    dp[idx]=temp;
    return temp;
}



ll dfs(ll src, ll p, vector<ll>& a, vector<pair<ll,ll> > adj[], vector<ll>& dp)
{
    ll can1=0;
    ll can2=0;
    vector<pair<ll,ll> >:: iterator it;
    for(it=adj[src].begin();it!=adj[src].end();it++)
    {
        if(((*it).first==p)) continue;
        ll st=max(0LL,dfs((*it).first,src,a,adj,dp)- (*it).second);
        if(st>can1)
        {
            can2=can1;
            can1=st;
        }
        else if(st>can2)
        {
            can2=st;
        }
    }
    dp[src]=a[src]+ can1+can2;
    return (max(can1,can2)+ a[src]);
}
/*  FINDING LCS IN (NLOGN) COMPLEXITY
int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	vector<int> d(n+1, 1000000000);
	for (int i = 0; i < n; i++) {
		*lower_bound(d.begin(), d.end(), a[i]) = a[i];
	}
	for (int i = 0; i <= n; i++) {
		if (d[i] == 1000000000) {
			cout << i << endl;
			return 0;
		}
	}
*/
struct comp1{
    bool operator ()(pair<ll,ll>& temp1, pair<ll,ll>& temp2)
{

}
};
bool comp2(pair<ll,ll>& temp1, pair<ll,ll>& temp2)
{
    return ((temp1.first<temp2.first) || ((temp1.first==temp2.first) && (temp1.second<temp2.second)));
}

ll getinv(ll temp)
{
    return power(temp,mod-2,mod);

}
void computeLPSArray(string pat, ll M, vector<ll>&lps)
{
    // length of the previous longest prefix suffix
    int len = 0;

    lps[0] = 0; // lps[0] is always 0

    // the loop calculates lps[i] for i = 1 to M-1
    ll i = 1;
    while (i < M) {
        if (pat[i] == pat[len]) {
            len++;
            lps[i] = len;
            i++;
        }
        else // (pat[i] != pat[len])
        {
            // This is tricky. Consider the example.
            // AAACAAAA and i = 7. The idea is similar
            // to search step.
            if (len != 0) {
                len = lps[len - 1];

                // Also, note that we do not increment
                // i here
            }
            else // if (len == 0)
            {
                lps[i] = 0;
                i++;
            }
        }
    }
}
ll fin(ll u, vector<ll>& parent)
{
    if(u==parent[u]) return u;
    else return parent[u]=fin(parent[u],parent);
}
void unio(ll u, ll v, vector<ll>& parent, vector<ll>& ran)
{
    u=fin(u,parent);
    v=fin(v,parent);
    if(u==v) return;
    if(ran[u]>=ran[v])
    {
        parent[v]=u;
        ran[u]+=ran[v];
    }
    else{
        parent[u]=v;
        ran[v]+=ran[u];
    }
}

void update(int x, int delta,vector<ll>& BIT )
{
    ll n=BIT.size()-1;
      for(; x <= n; x += x&-x)
        BIT[x] += delta;
}
int query(int x, vector<ll>& BIT)
{
     int sum = 0;
     for(; x > 0; x -= x&-x)
        sum += BIT[x];
     return sum;
}
ll k;
string s;
ll dp[2000][2000];
ll recur(ll idx, ll match, ll n)
{
    if(idx==(n+k))
    {
        if(match==n) return 1;
        else return 0;
    }
    if(dp[idx][match]!=-1) return dp[idx][match];
    ll temp=0;
    if(match<n)
    {
    ll d=s[match]-'a';
    for(ll i=0;i<26;i++)
    {
        if(i==d)
        {
            temp=add(temp,recur(idx+1,match+1,n));
        }
        else
            temp=add(temp,recur(idx+1,match,n));
    }
    }
    else
    {
        for(ll i=0;i<26;i++)
    {
            temp=add(temp,recur(idx+1,match,n));
    }
    }
    dp[idx][match]=temp;
    return temp;
}
int main()
{
    //ll k;
    //string s;
    cin>>k>>s;
    ll n=s.length();
    memset(dp,-1,sizeof(dp));
    ll ans=recur(0,0,n);
    cout<<ans;
}