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
A = input().split()
n = int(A[0])
k = int(A[1])
fact = [1]*(n+1)
for i in range (1,n+1):
    fact[i] = fact[i-1]*(i)
der = [1]*(n+1)
der[0] = 1
der[1] = 0
for i in range (2,n+1):
    der[i] = (i-1)*(der[i-1]+der[i-2])
def comb(n,k):
    return(fact[n]//(fact[k]*fact[n-k]))
tot = 0
for i in range (0,k+1):
    tot = tot + comb(n,i)*der[i]
print(tot)