/*
   題意：
       dp[n,k] 表示n個數有k個ai>i
       考慮第n個數
       數n與前面任何一個ai<=i的交換，E-value增大1
       數n與前面任何一個ai>i的交換或者an = n，E-value不變
       dp[n,k] = (n-1-(k-1))*dp[n-1,k-1] + (k+1)*dp[n-1,k]
               = (n-k)*dp[n-1,k-1] + (k+1)*dp[n-1,k]
 */
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

const int MOD = 1000000007;

int dp[1010][1010];

void init()
{
    dp[0][0] = 1;
    for(int n=1;n<=1000;n++)
        for(int k=0;k<=n;k++)
        {
            dp[n][k] = ((long long)(k+1)*dp[n-1][k])%MOD;
            if(k)
                dp[n][k] = (dp[n][k] + (long long)(n-k)*dp[n-1][k-1])%MOD;
        }
}
int main()
{
   // freopen("in","r",stdin);
    init();
    int N,K;
    while(~scanf("%d%d",&N,&K))
    {
        printf("%d\n",dp[N][K]);
    }
    return 0;
}