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 sys
input = sys.stdin.readline
 
############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
def fact(n):
    if n==0:
        return(1)
    p=1
    for i in range(1,n+1):
        p=p*i
    return(p)
def de(n):
    o=0
    a=fact(n)
    for i in range(n+1):
        o=o + (a*((-1)**i))/fact(i)
    return(o)
n,k=invr()
out=0
for i in range(n-k,n+1):
    out=out+(fact(n)/(fact(i)*fact(n-i)))*de(n-i)
print(int(out))
