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 \(p_i = i\).

Your task is to count the number of almost identity permutations for given numbers \(n\) and \(k\).

Input

The first line contains two integers \(n\) and \(k\) \((4 ≤ n ≤ 1000, 1 ≤ k ≤ 4)\).

Output

Print the number of almost identity permutations for given \(n\) and \(k\).

Examples

Input

4 1

Output

1

Input

4 2

Output

7

Input

5 3

Output

31

Input

5 4

Output

76

题意

给出\(n\)的全排列，求有多少种排列，满足至少\(n-k\)个位置上的数和下标相同（下标从\(1\)开始）

思路

因为\(1\leq k\leq 4\)，所以可以将题意转换一下：在\(n\)的全排列中，找到\(k\)个数，数和下标的值全都不相等

我们可以从\(n\)个数中，随机选出\(k\)个数，让这\(k\)个数全都没有放在正确的位置上，选\(k\)个数，我们可以用组合数来求，然后用错排公式来求有多少个数没放在正确的位置上。

因为\(k\)只有四个值，直接计算错排公式的值即可

最后将\(1\)～\(k\)中的这些值加起来即可

代码

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
ll C(int n,int m)
{
    ll fenmu=1LL;
    ll fenzi=1LL;
    for(int i=1;i<=m;i++)
    {
        fenmu=1LL*fenmu*(n-i+1);
        fenzi=1LL*fenzi*i;
    }
    return fenmu/fenzi;
}
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    ll ans=0;
    if(k>=1)
        ans+=1;
    if(k>=2)
        ans+=(n*(n-1)/2);
    if(k>=3)
        ans+=2*C(n,3);
    if(k>=4)
        ans+=9*C(n,4);
    cout<<ans<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}