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 Rust (1.42.0) (judge verdict: Accepted)

use std::cmp::*;
use std::io::*;

const MOD: u64 = 1_000_000_007;

mod enumeration {
    pub struct Enumeration {
        fact: Vec<u64>,
        finv: Vec<u64>,
        modulo: u64,
    }
    impl Enumeration {
        pub fn new(n: usize, modulo: u64) -> Enumeration {
            let mut fact: Vec<u64> = vec![1; n];
            let mut inv: Vec<u64> = vec![1; n];
            let mut finv: Vec<u64> = vec![1; n];
            for i in 2..n {
                fact[i] = fact[i - 1] * (i as u64) % modulo;
                inv[i] = modulo - (modulo / (i as u64)) * inv[(modulo as usize % i)] % modulo;
                finv[i] = inv[i] * finv[i - 1] % modulo;
            }
            Enumeration {
                fact: fact,
                finv: finv,
                modulo: modulo,
            }
        }
        pub fn c(&self, n: usize, k: usize) -> u64 {
            if n < k {
                return 0;
            }
            self.fact[n] * self.finv[n - k] % self.modulo * self.finv[k] % self.modulo
        }
    }
}

fn modpow(mut x: u64, mut e: u64) -> u64 {
    let mut res = 1;
    while e > 0 {
        if e & 1 == 1 {
            res = x * res % MOD;
        }
        x = x * x % MOD;
        e >>= 1;
    }
    res
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let k: usize = itr.next().unwrap().parse().unwrap();
    let s: Vec<char> = itr.next().unwrap().chars().collect();
    let n = s.len();

    let unko = enumeration::Enumeration::new(n + k + 1, MOD);
    let mut ans = 0;
    for i in n..k + n {
        ans = (ans + unko.c(k + n, i) * modpow(25, (k + n - i) as u64)) % MOD;
    }
    println!("{}", (ans + 1) % MOD);
}