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
from math import factorial
def comb(n, r):
    if(r ==1 ):
        return n
    elif(r == 2):
        return (n * (n - 1)) // 2
    elif(r == 3):
        return (n * (n -1) * (n - 2))//6
    else:
        return (n * (n- 1) * (n-2) * (n-3))//24
def derange(n):
    if(n == 1):
        return 0
    elif(n == 2):
        return 1
    elif(n == 3):
        return 2
    elif(n == 4):
        return 9
n,k = map(int,input().split())

ans = 0
for i in range(1, k+1):
    ans += comb(n, i) * int(derange(i))
print(ans+1)