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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.net.ConnectException;
import java.rmi.dgc.Lease;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Deque;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.Objects;
import java.util.Set;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

import static java.util.Comparator.*;

public class Main {
	public static void main(String[] args) {
		Main main = new Main();
		main.solve();
		main.out.close();
	}
	// ======================================================================
	public void solve() {
		int K = ni();
		int N = ns().length();
		long ans = 0, wk;
		COMinit();
		for (int i = 0; i <= K; i++) {
			wk = modpow(26, K-i);
			wk = (wk * modpow(25, i)) % mod;
			wk = (wk * COM(i + N - 1, i)) % mod;
			ans = (ans + wk) % mod;
		}
		out.println(ans);
	}
	// a の p 乗を求めるアルゴリズム				
	// p = 30 の時、p の変化は　30 -> 15 -> 14 -> 7 -> 6 -> 3 -> 2 -> 1 -> 0				
	// と、半分半分になっていくので、計算量は O(log p) となる				
	int mod = (int)(1e9) + 7;				
	long modpow(long a, int p) {				
		if(p == 0)	return 1;		
		if(p % 2 == 0) {			
			int halfP = p / 2;                        // 偶数なら		
			long half = modpow(a, halfP);    // 乗数を半分にする		
			return half * half % mod;           // 計算結果を２回かける　→　乗数を戻す		
		} else {			
			return a * modpow(a, p - 1) % mod; // 乗数から１を引いた結果に a を掛ける		
		}			
	}				
	int MAX = 2100000;
	int MOD = (int)(1e9) + 7;;
	long[] fac, finv, inv;			// a! (fac[a])   (a!)の-1乗 (finv[a])
	
	// テーブルを作る前処理　→　これを呼んでから COM() を呼ぶ
	void COMinit() {
		fac = new long[MAX];
		finv = new long[MAX];
		inv = new long[MAX];
		fac[0] = fac[1] = 1;
		finv[0] = finv[1] = 1;
		inv[1] = 1;
		for (int i = 2; i < MAX; i++){
			fac[i] = fac[i - 1] * i % MOD;
			inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
			finv[i] = finv[i - 1] * inv[i] % MOD;
		}
	}
	// 組み合わせの数を計算する(a個の中からb個を選ぶ）
	long COM(int n, int k){
		if (n < k)		return 0;
		if (n < 0 || k < 0)		return 0;
		return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
	}
	
	// ------------------------------------------
	// ライブラリ
	// ------------------------------------------
	// Print
	private PrintWriter out = new PrintWriter(System.out);
	
	// Scanner
	private FastScanner scan = new FastScanner();
	int ni(){ return scan.nextInt();}
	int[] ni(int n){int[] a = new int[n]; for(int i = 0; i < n; i++){a[i] = ni();} return a;}
	int[][] ni(int y, int x){int[][] a = new int[y][x];
	for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = ni();}} return a;}
	long nl(){return scan.nextLong();}
	long[] nl(int n){long[] a = new long[n]; for(int i = 0; i < n; i++){a[i] = nl();} return a;}
	long[][] nl(int y, int x){long[][] a = new long[y][x];
	for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = nl();}} return a;}
	String ns(){return scan.next();}
	String[] ns(int n){String[] a = new String[n]; for(int i = 0; i < n; i++){a[i] = ns();} return a;}
	String[][] ns(int y, int x){String[][] a = new String[y][x];
	for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = ns();}} return a;}
}
class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;
	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		}else{
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}
	private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
	private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
	public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
	public String next() {
		if (!hasNext()) throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while(isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	public long nextLong() {
		if (!hasNext()) throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while(true){
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			}else if(b == -1 || !isPrintableChar(b)){
				return minus ? -n : n;
			}else{
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
		return (int) nl;
	}
	public double nextDouble() { return Double.parseDouble(next());}
}