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
import operator as op
from functools import reduce
def ncr(n, r):
    r = min(r, n-r)
    numer = reduce(op.mul, range(n, n-r, -1), 1)
    denom = reduce(op.mul, range(1, r+1), 1)
    return numer / denom
def countDer(n): 
    der = [0 for i in range(n + 10)] 
    der[0] = 1
    der[1] = 0
    der[2] = 1
    for i in range(3, n + 1): 
        der[i] = (i - 1) * (der[i - 1] + der[i - 2]) 
    return der[n] 
n, k = map(int, input().split())
ans = 0
for i in range(n - k, n + 1):
    ans += ncr(n, i) * countDer(n - i)
print(int(ans))
		       	 		 			 	   			 	  	