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: Time Limit Exceeded)

#![allow(unused_imports)]
#![allow(non_snake_case, unused)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;

// https://atcoder.jp/contests/hokudai-hitachi2019-1/submissions/10518254

macro_rules! eprint {
    ($($t:tt)*) => {{
        use ::std::io::Write;
        let _ = write!(::std::io::stderr(), $($t)*);
    }};
}
macro_rules! eprintln {
    () => { eprintln!(""); };
    ($($t:tt)*) => {{
        use ::std::io::Write;
        let _ = writeln!(::std::io::stderr(), $($t)*);
    }};
}
macro_rules! dbg {
    ($v:expr) => {{
        let val = $v;
        eprintln!("[{}:{}] {} = {:?}", file!(), line!(), stringify!($v), val);
        val
    }}
}

macro_rules! mat {
    ($($e:expr),*) => { Vec::from(vec![$($e),*]) };
    ($($e:expr,)*) => { Vec::from(vec![$($e),*]) };
    ($e:expr; $d:expr) => { Vec::from(vec![$e; $d]) };
    ($e:expr; $d:expr $(; $ds:expr)+) => { Vec::from(vec![mat![$e $(; $ds)*]; $d]) };
}

macro_rules! ok {
    ($a:ident$([$i:expr])*.$f:ident()$(@$t:ident)*) => {
        $a$([$i])*.$f($($t),*)
    };
    ($a:ident$([$i:expr])*.$f:ident($e:expr$(,$es:expr)*)$(@$t:ident)*) => { {
        let t = $e;
        ok!($a$([$i])*.$f($($es),*)$(@$t)*@t)
    } };
}

pub fn readln() -> String {
    let mut line = String::new();
    ::std::io::stdin().read_line(&mut line).unwrap_or_else(|e| panic!("{}", e));
    line
}

macro_rules! read {
    ($($t:tt),*; $n:expr) => {{
        let stdin = ::std::io::stdin();
        let ret = ::std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
            let line = line.unwrap();
            let mut it = line.split_whitespace();
            _read!(it; $($t),*)
        }).collect::<Vec<_>>();
        ret
    }};
    ($($t:tt),*) => {{
        let line = readln();
        let mut it = line.split_whitespace();
        _read!(it; $($t),*)
    }};
}

macro_rules! _read {
    ($it:ident; [char]) => {
        _read!($it; String).chars().collect::<Vec<_>>()
    };
    ($it:ident; [u8]) => {
        Vec::from(_read!($it; String).into_bytes())
    };
    ($it:ident; usize1) => {
        $it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
    };
    ($it:ident; [usize1]) => {
        $it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
    };
    ($it:ident; [$t:ty]) => {
        $it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
    };
    ($it:ident; $t:ty) => {
        $it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
    };
    ($it:ident; $($t:tt),+) => {
        ($(_read!($it; $t)),*)
    };
}


pub fn main() {
    let _ = ::std::thread::Builder::new().name("run".to_string()).stack_size(32 * 1024 * 1024).spawn(run).unwrap().join();
}

const MOD: i64 = 1_000_000_007;
const INF: i64 = std::i64::MAX/3;

fn solve() {
    let k = read!(usize);
    let s = read!([char]);
    let n = s.len();
    let mut dp = vec![vec![0;n+1];2];
    dp[0][0] = 1;
    let mut idx = 0;
    for i in 0..n+k {
        for j in 0..n+1 {
            dp[1-idx][j] = dp[idx][j]*25;
            dp[1-idx][j] %= MOD;
            if j>0 {
                dp[1-idx][j] += dp[idx][j-1];
                dp[1-idx][j] %= MOD;
                if j==n {
                    dp[1-idx][j] += dp[idx][j];
                    dp[1-idx][j] %= MOD;
                }
            }
        }
        // for j in 0..n+1 {
        //     print!("{} ",dp[1-idx][j]);
        // }
        // println!();
        idx = 1 - idx;
    }
    println!("{}",dp[idx][n]);
}

fn run() {
    solve();
}