A permutation *p* of size *n* is an array such that every integer from 1 to *n* occurs exactly once in this array.

Let's call a permutation an almost identity permutation iff there exist at least *n* - *k* indices *i* (1 ≤ *i* ≤ *n*) such that *p*<sub class="lower-index">*i*</sub> = *i*.

Your task is to count the number of almost identity permutations for given numbers *n* and *k*.
The first line contains two integers *n* and *k* (4 ≤ *n* ≤ 1000, 1 ≤ *k* ≤ 4).
none
4 1
1
4 2
7
5 3
31
5 4
76
200 1
1
200 2
19901
200 3
2646701
200 4
584811251
400 1
1
400 2
79801
400 3
21253401
400 4
9477912501
600 1
1
600 2
179701
600 3
71820101
600 4
48187303751
800 1
1
800 2
319601
800 3
170346801
800 4
152620985001
1000 1
1
1000 2
499501
1000 3
332833501
1000 4
373086956251
4 4
24
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#  code

from itertools import permutations as perm
from math import factorial as fac
def calc(n):
    a = [i + 1 for i in range(n)]
    cnt = 0
    for i in perm(a):
        ok = True
        for idx,j in enumerate(i):
            if idx + 1 == j:
                ok = False
        cnt += 1 if ok else 0
    return cnt


def main():
    c = [calc(i + 1) for i in range(4)]
    print(c)

    n,k = map(int , input().split())
    fa = 1
    ans = 0

    for i in range(k):
        fa = fa * (n - i)

    while k > 0:
        if k == 1:
            ans = ans + 1
        else:
            ans = ans + (fa // fac(k)) * c[k - 1]
            # print((fa // fac(k)))
        k -= 1
        fa //= (n - k)
    print(ans)
    return

if __name__ == "__main__":
    main()