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
def countDer(n):

    der = [0 for i in range(max(3,(n + 1)))]

    # Base cases
    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]


def nCr(n, r):
    return (fact(n) / (fact(r)
                       * fact(n - r)))


# Returns factorial of n
def fact(n):
    res = 1

    for i in range(2, n + 1):
        res = res * i

    return res
n,k=map(int,input().split())
j=(n-k)
s=0
while(j<=n):
    s+=(nCr(n,j)*(countDer(n-j)))
    j+=1
print(int(s))
