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 math
from itertools import permutations
def ncr(n,r):
    return (math.factorial(n)//(math.factorial(r)*math.factorial(n-r)))

def left_one(left):
    temp=[x for x in range(left)]
    cnt=0
    allperm=list(permutations(temp))
    for x in allperm:
        chk=0
        for z in range(len(x)):
            if x[z]==z:
                chk=1
                break
        if chk==0:
            #print(x)
            cnt+=1
    return cnt
n,k=map(int,input().split(" "))
k=n-k
ans=1
for x in range(k,n):
    #print(left_one(n-x),ncr(n,x))
   # print(n-x)
    ans+=ncr(n,x)*left_one(n-x)
print(ans)
