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
fact = [0] * 1001
fact[0] = 1
for i in range(1, 1001):
    fact[i] = fact[i - 1] * i
n, k = map(int, input().split())
mess = [0] * 1001

def get_mess(x, fact):
    ans = 0
    for i in range(0, x + 1):
        sign = 1 - 2 * (i % 2)
        ans = ans + sign * (fact[x] // fact[i])
    return ans
m = n - k
ans = 0
for i in range(m, n + 1):
    ans += get_mess(n - i, fact) * (fact[n] // fact[i] // fact[n - i])
print(ans)

    
    

# Tue Nov 09 2021 16:29:46 GMT+0000 (Coordinated Universal Time)
