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 PyPy2 (7.3.0) (judge verdict: Accepted)

FAST_IO = 0
if FAST_IO:
    import io, sys, atexit
    rr = iter(sys.stdin.read().splitlines()).next
    sys.stdout = _OUTPUT_BUFFER = io.BytesIO()
    @atexit.register
    def write():
        sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
else:
    rr = raw_input
rri = lambda: int(rr())
rrm = lambda: map(int, rr().split())
rrmm = lambda n: [rrm() for _ in xrange(n)]

####

def solve(K, S):
    MOD = 10 ** 9 + 7
    N = len(S)
    N2 = N + K

    BIG =N2
    fac = [1, 1]
    for x in xrange(2, BIG+1):
        fac.append(fac[-1] * x  %MOD)
    inv = [1] * (BIG + 1)
    inv[BIG] = pow(fac[-1], MOD-2, MOD)
    for x in xrange(BIG-1, -1, -1):
        inv[x] = inv[x+1] * (x+1) % MOD

    def binom(n, k):
        ans = fac[n]
        ans *= inv[n-k]
        ans %= MOD
        
        ans *= inv[k]
        ans %= MOD
        return ans

    pow25 = [1]
    for x in xrange(1, BIG + 1):
        pow25.append(pow25[-1] * 25  %MOD)

    ans = 0
    for n in xrange(N, N2 + 1):
        ans += binom(N2, n)   * pow25[N2- n] % MOD
        ans %= MOD
    return ans

print solve(rri(), rr())