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

n = int(input())
k = int(input())

def factorial(n):
  if n == 0 or n == 1:
    return 1
  else:
    return n * factorial(n - 1)


def binomial(n, k):
  return factorial(n) // (factorial(k) * factorial(n - k))
if k == 1:
  print(1)
elif k == 2:
  print(n * (n - 1) // 2 + 1)
elif k == 3:
  print(n * (n - 1) * (n - 2) // 3 + n * (n - 1) // 2 + 1)
elif k == 4:
  print(binomial(n, k) * 9 + n * (n - 1) * (n - 2) // 3 + n * (n - 1) // 2 + 1)
else:
  print("Invalid value of k")