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



Sample input:
5
oof


Sample output:
575111451

Submission status: Time Limit Exceeded

Solution code:
k = int(input())
s = input()
mod = 10 ** 9 + 7

nagasa = len(s)

n = nagasa + k

N = n + 5
bikkuri = [0 for i in range(N)]
bikkuri[0] = 1
gyaku = [0 for i in range(N)]
gyaku[0] = 1
for i in range(1, N):
    bikkuri[i] = (i * bikkuri[i - 1]) % mod
    gyaku[i] = pow(bikkuri[i], mod - 2, mod)


def comb(n, r):
    if r < 0 or n - r < 0:
        return 0
    return (bikkuri[n] * gyaku[r] * gyaku[n - r]) % mod


ans = 0
for i in range(k + 1):
    temp = 0
    temp += comb(nagasa - 1 + i, nagasa - 1) * pow(25, i, mod)
    temp %= mod
    temp *= pow(26, k - i, mod)
    temp %= mod
    ans += temp
    ans %= mod

print(ans)

Submission status: Time Limit Exceeded

Solution code:
mod = 10**9 + 7

fact_range = 10**6 * 2
facts = [1] * (fact_range + 1)
for i in range(0, fact_range):
    facts[i+1] = facts[i] * (i + 1) % mod

ifacts = [1] * (fact_range + 1)
ifacts[fact_range] = pow(facts[fact_range], mod - 2, mod)
for i in range(fact_range, 0, -1):
    ifacts[i-1] = ifacts[i] * i % mod

def comb(n, k):
    if k < 0 or n < k:
        return 0
    else:
        return facts[n] * ifacts[n-k] % mod * ifacts[k] % mod

k = int(input())
s = len(input())
ans = 0
for a in range(k + 1):
    ans += pow(25, a, mod) * pow(26, k-a, mod) % mod * comb(s-1+a, a)
    ans %= mod
print(ans)

Submission status: Time Limit Exceeded

Solution code:
import sys

stdin = sys.stdin


def ni(): return int(ns())


def na(): return list(map(int, stdin.readline().split()))


def naa(N): return [na() for _ in range(N)]


def ns(): return stdin.readline().rstrip()  # ignore trailing spaces

# 大きい数用


def modPow(a, n, p):
    if n == 0:
        return 1
    if n == 1:
        return a
    if n % 2 == 1:
        return (a * (modPow(a, n // 2, p) ** 2)) % p
    else:
        return (modPow(a, n // 2, p) ** 2) % p


def modInverse(a, p):
    # calculates the modular multiplicative of a mod m.
    # (assuming p is prime).
    return modPow(a, p - 2, p)


def modBinomial(n, k, p):
    # calculates C(n,k) mod p (assuming p is prime).

    numerator = 1  # n * (n-1) * ... * (n-k+1)
    for i in range(k):
        numerator = (numerator * (n - i)) % p

    denominator = 1  # k!
    for i in range(1, k + 1):
        denominator = (denominator * i) % p

    # numerator / denominator mod p.
    return (numerator * modInverse(denominator, p)) % p


K = ni()
S = ns()
s_len = len(S)
mod = 10 ** 9 + 7
ans = 0

total = K + s_len

rui = 1

cmb = 1

for i in range(total, s_len - 1, -1):
    ans = (ans + cmb * rui) % mod
    # print(ans)
    rui = (rui * 25) % mod
    cmb = (cmb * i * modInverse(total - i + 1, mod)) % mod
print(ans)


# a_array = [[0] * 9 for _ in range(9)]
# a_array[0][0] = 1
# for i in range(1,9):
#     for j in range(9):
#         if j == 0:
#             a_array[i][j] = a_array[i-1][j] * 25
#         else:
#             a_array[i][j] = a_array[i-1][j] * 25 + a_array[i-1][j-1]

# print(sum(a_array[-1][3:]) % (10 ** 9 + 7))

Submission status: Time Limit Exceeded

Solution code:
MOD = 10**9+7
MAXN = 2*10**6+5
f = [1]
for i in range(MAXN):
    f.append(f[-1] * (i+1) % MOD)
def nCr(n, r, mod=MOD):
    return f[n] * pow(f[r], mod-2, mod) * pow(f[n-r], mod-2, mod) % mod

K = int(input())
S = input()
R = len(S)
N = K + R
ans = 0
for r in range(R,N+1):
    power = pow(25, N-r, MOD)
    ans += nCr(N, r) * power % MOD
    ans %= MOD
print(ans)

Submission status: Time Limit Exceeded

Solution code:
def prepare(n, MOD):
    f = 1
    factorials = [1]
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    inv = pow(f, MOD - 2, MOD)
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
    return factorials, invs


k = int(input())
s = input()
l = len(s)
MOD = 10 ** 9 + 7

facts, finvs = prepare(k + l, MOD)

ans = 0
for i in range(k + 1):
    r = k - i
    pat = facts[l + r - 1] * finvs[l - 1] * finvs[r] % MOD
    ans = (ans + pat * pow(25, r, MOD) * pow(26, i, MOD)) % MOD
print(ans)

Submission status: Time Limit Exceeded

Solution code:
import sys
sys.setrecursionlimit(int(1e6 + 3))

K = int(input())
lenS = len(input())
mod = int(1e9 + 7)

def fpow(x, n):
    ret = 1
    tmp = x
    while n:
        if n & 1:
            ret = ret * tmp % mod
        tmp = tmp * tmp % mod
        n >>= 1
    return ret
    
def memorize(f):
    dic = {}
    def main(i, j):
        if (i, j) not in dic:
            dic[(i, j)] = f(i, j)
        return dic[(i, j)]
    return main
        
@memorize
def dp(now, posK):
    if posK == 0:
        return fpow(26, now)
    if now == posK:
        return 1
    elif now < posK:
        return 0
    else:
        return (25 * dp(now - 1, posK) % mod + dp(now - 1, posK - 1) % mod) % mod

dp2 = {}
for i in range(1, K + lenS + 1):
    for j in range(lenS + 1):
        if j == i:
            dp2[(i, j)] = 1
            break
        elif j == 0:
            dp2[(i, j)] = fpow(26, i)
        else:
            dp2[(i, j)] = (dp2[(i - 1, j)] * 25 + dp2[(i - 1, j - 1)]) % mod
    
print(dp2[(K + lenS, lenS)])

Submission status: Time Limit Exceeded

Solution code:
K=int(input())
S=input()
mod=10**9+7

FACT=[1]
for i in range(1,2*10**6+3):
    FACT.append(FACT[-1]*i%mod)

FACT_INV=[pow(FACT[-1],mod-2,mod)]
for i in range(2*10**6+2,0,-1):
    FACT_INV.append(FACT_INV[-1]*i%mod)

FACT_INV.reverse()

def Combi(a,b):
    if 0<=b<=a:
        return FACT[a]*FACT_INV[b]*FACT_INV[a-b]%mod
    else:
        return 0


LEN=len(S)
ANS=0
for i in range(K+1):
    ANS+=Combi(i+LEN-1,LEN-1)*pow(25,i,mod)*pow(26,K-i,mod)
print(ANS%mod)

Submission status: Time Limit Exceeded

Solution code:
MOD = 10 ** 9 + 7


def solve(k, s):
    res = 0
    factorial = [1] * (2 * 10 ** 6)
    factorial_inv = [1] * (2 * 10 ** 6)
    for i in range(1, 2 * 10 ** 6):
        factorial[i] = factorial[i - 1] * i % MOD
    factorial_inv[-1] = pow(factorial[-1], MOD - 2, MOD)
    for i in range(2 * 10 ** 6 - 1, 0, -1):
        factorial_inv[i - 1] = factorial_inv[i] * i % MOD
    # print(factorial[:5])
    # print(factorial_inv[:5])
    for i in range(k + 1):
        left = k - i + len(s) - 1
        right = len(s) - 1
        ncr = (factorial[left] * factorial_inv[right] * factorial_inv[left - right]) % MOD
        add = (ncr * pow(25, k - i, MOD) * pow(26, i, MOD)) % MOD
        res += add
        res %= MOD
    # print(res)
    return res


def main():
    k = int(input())
    s = input()
    res = solve(k, s)
    print(res)


def test():
    assert solve(1, "abc") == 25 + 25 + 25 + 26
    assert solve(5, "oof") == 575111451
    assert solve(37564, "whydidyoudesertme") == 318008117


if __name__ == "__main__":
    test()
    main()

Submission status: Time Limit Exceeded

Solution code:
SIZE=2*10**6; MOD=10**9+7 #998244353 #ここを変更する

SIZE += 1
inv = [0]*SIZE  # inv[j] = j^{-1} mod MOD
fac = [0]*SIZE  # fac[j] = j! mod MOD
finv = [0]*SIZE # finv[j] = (j!)^{-1} mod MOD
fac[0] = fac[1] = 1
finv[0] = finv[1] = 1
for i in range(2,SIZE):
    fac[i] = fac[i-1]*i%MOD
finv[-1] = pow(fac[-1],MOD-2,MOD)
for i in range(SIZE-1,0,-1):
    finv[i-1] = finv[i]*i%MOD
    inv[i] = finv[i]*fac[i-1]%MOD

def choose(n,r): # nCk mod MOD の計算
    if 0 <= r <= n:
        return (fac[n]*finv[r]%MOD)*finv[n-r]%MOD
    else:
        return 0



# coding: utf-8
# Your code here!
import sys
read = sys.stdin.read
readline = sys.stdin.readline

k, = map(int,readline().split())
s = input()

MOD = 10**9+7

l = len(s)
ans = pow(26,l+k,MOD)
for p in range(l):
    ans -= pow(25,k+l-p)*choose(l+k,p)

print(ans%MOD)

Submission status: Accepted

Solution code:
# from sys import stdin
# input = stdin.readline
from collections import Counter

MOD = 10**9 + 7
MAX = 2000001
def solve():
    k = int(input())
    s = input().strip()
    ns = len(s)
    res = 0
    fac = [1,1] + [0] * MAX
    inv = [1,1] + [0] * MAX
    for i in range(2,MAX+1):
        fac[i] = fac[i-1] * i % MOD
    inv[MAX] = pow(fac[MAX],MOD-2,MOD)
    for i in range(MAX-1,1,-1):
        inv[i] = inv[i+1]*(i+1)%MOD

    def c(a,b):
        return fac[b]*inv[a]*inv[b-a] % MOD



    for left in range(0,k+1):
        right = k - left
        temp = pow(25,left,MOD) * pow(26,right,MOD)
        temp *= c(ns - 1, left + ns -1)
        temp %= MOD
        res += temp
        res %= MOD
    print(res)







if __name__ == '__main__':
    solve()

Submission status: Accepted

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product, combinations_with_replacement # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd
    def gcd(a,b):
        # from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
        # => 最近の環境でfractions.gcdを使うとTLEする。
        if a<b:
            a, b = b, a
        while a%b !=0 :
            a, b = b, a%b
        return b

    # def gcd(p,q):
    #     while q != 0:
    #         r = p % q
    #         p = q
    #         q = r
    #     return p

    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)

    # Eratosthenesの篩により、整数end未満の素数を列挙する
    def sieve_of_eratosthenes(end, typecode="L"):
        # len(sieve_of_eratosthenes(10**7)) => 664579; 514ms in pypy
        # len(sieve_of_eratosthenes(10**6)) => 78498; 134ms in pypy

        # https://hamukichi.hatenablog.jp/entry/2016/02/17/215948#Eratosthenes%E3%81%AE%E7%AF%A9%E3%81%A8%E5%8C%BA%E9%96%93%E7%AF%A9
        import array
        import textwrap
        assert end > 1
        # 整数iが素数であるかをis_prime[i]が示す
        # 最初はすべてTrueで初期化しておく
        # 最終的にprimesではなくこれを返してもよい
        is_prime = array.array("B", (True for i in range(end)))
        # 0, 1はいずれも素数ではない
        is_prime[0] = False
        is_prime[1] = False
        # 素数を格納する配列
        primes = array.array(typecode)
        # 篩う
        for i in range(2, end):
            if is_prime[i]:
                primes.append(i)
                for j in range(2 * i, end, i):
                    is_prime[j] = False  # 素数ではないため除外する
        return primes

    class PrimeFactorization:
        '''初期化にO(NloglogN)/ 因数分解クエリをO(log(k))/ 素数判定クエリをO(1)
        >>> pf = PrimeFactorization(100)
        >>> pf.query(100)
        ((2, 2), (5, 2))
        
        '''
        # 自分で実装した。
        # https://atcoder.jp/contests/abc177/editorial/82
        # 高速素因数分解
        # 問題： A 以下の数が N 個与えられる。全て素因数分解せよ。
        # 前計算としてエラトステネスの篩を行い、「その数をふるい落とした素数」を配列 D に記録します。
        # 例えば D[4]=D[6]=2,D[35]=5 です。x が素数のときは D[x]=x としておきます。この配列はエラトステネスの篩と同様 O(AloglogA) で構築できます。
        # D[x] は x を割り切る最小の素数なので、この配列 D を利用すると素因数分解を行うときに「試し割り」をする必要がなくなり(D[x]で割ればよい)、1つの数の素因数分解が素因数の個数である O(logA) でできるようになります。
        def __init__(self, Num=10**6):
            from collections import defaultdict
            from math import floor
            self.Num = Num
            self.D = [-1] * (Num+1)
            for i in range(2, Num+1):
                if self.D[i] != -1:
                    continue
                self.D[i] = i
                for j in range(1, floor(Num/i)+1):
                    if self.D[j*i] != -1:
                        continue
                    else:
                        self.D[j*i] = i
        
        def query(self, k):
            ans = defaultdict(int)
            tmp = k
            while self.D[tmp] != -1:
                div = self.D[tmp]
                ans[div] += 1
                tmp = tmp//div
            return tuple(ans.items())
        
        def is_prime(self, k):
            return k == self.D[k]

    # first create factorial_list
    # fac_list = modfactorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrstuvwxyz') # 26
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something like:
        >>      pass
        >>  else:
        >>      do something like:
        >>      L[find_root(a)] = L[find_root(a)] + L[find_root(b)]
        >>      L[find_root(b)] = L[find_root(a)]
        >>      
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # c++ set
    # https://nagiss.hateblo.jp/entry/2020/09/08/203701
    
    class SegTree:
        """
        ref: https://qiita.com/takayg1/items/c811bd07c21923d7ec69
        init(init_val, ide_ele): 配列init_valで初期化 O(N)
        update(k, x): k番目の値をxに更新 O(logN)
        query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
        
        ex: seg = SegTree([14, 5, 9, 13], min, float('inf'))
        """
        def __init__(self, init_val, segfunc, ide_ele):
            """
            init_val: 配列の初期値
            segfunc: 区間にしたい操作
            ide_ele: 単位元
            n: 要素数
            num: n以上の最小の2のべき乗
            tree: セグメント木(1-index)
            """
            n = len(init_val)
            self.segfunc = segfunc
            self.ide_ele = ide_ele
            self.num = 1 << (n - 1).bit_length()
            self.tree = [ide_ele] * 2 * self.num
            # 配列の値を葉にセット
            for i in range(n):
                self.tree[self.num + i] = init_val[i]
            # 構築していく
            for i in range(self.num - 1, 0, -1):
                self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

        def update(self, k, x):
            """
            k番目の値をxに更新
            k: index(0-index)
            x: update value
            """
            k += self.num
            self.tree[k] = x
            while k > 1:
                self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
                k >>= 1

        def query(self, l, r):
            """
            [l, r)のsegfuncしたものを得る
            l: index(0-index)
            r: index(0-index)
            """
            res = self.ide_ele
            l += self.num
            r += self.num
            while l < r:
                if l & 1:
                    res = self.segfunc(res, self.tree[l])
                    l += 1
                if r & 1:
                    res = self.segfunc(res, self.tree[r - 1])
                l >>= 1
                r >>= 1
            return res

    # ************ Graph ************
    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 最短経路問題: https://ja.wikipedia.org/wiki/%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E5%95%8F%E9%A1%8C
    
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    # Warshall Floyd O(V**3)
    def warshall_floyd(number_vertex, XYD, shift_one_for_vertex=False):
        # ref: https://juppy.hatenablog.com/entry/2018/11/01/%E8%9F%BB%E6%9C%AC_python_%E5%85%A8%E7%82%B9%E5%AF%BE%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E6%B3%95%EF%BC%88%E3%83%AF%E3%83%BC%E3%82%B7%E3%83%A3%E3%83%AB%E3%83%95%E3%83%AD%E3%82%A4%E3%83%89%E6%B3%95
        # number_vertex: len(vertex)
        # XYD: list of distance from x to y
        # shift_one_for_vertex: XYDの値が1~Nの時(0~N-1でない時)に調整するフラグ。
        # d[i][j]: iからjへの最短距離
        d = [[float("inf")]*number_vertex for i in range(number_vertex)]
        for x, y, dist in XYD:
            if shift_one_for_vertex:
                x = x - 1
                y = y - 1
            d[x][y] = dist
            d[y][x] = dist
        for i in range(number_vertex):
            d[i][i] = 0 # distance of same vertex is 0
        for k in range(number_vertex):
            for i in range(number_vertex):
                for j in range(number_vertex):
                    d[i][j] = min(d[i][j],d[i][k] + d[k][j])
        return d


    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    #%%
    K = r_int()
    S = len(r())
    
    ans = 0
    MOD = 10**9+7
    l = [1] * (K+1)
    r = [1] * (K+1)
    m = [1] * (K+1)
    for i in range(1, K+1):
        l[i] = (l[i-1] * 25)%MOD
        r[i] = (r[i-1] * 26)%MOD
        m[i] = (moddiv(m[i-1] * (S-1+i) , (i)))%MOD
    
    for i in range(K+1):
        ans = (ans + m[i]*l[i]*r[K-i])%MOD
    
    print(int(ans))
    
    #%%
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()