Little penguin Polo loves his home village. The village has n houses, indexed by integers from 1 to n. Each house has a plaque containing an integer, the i-th house has a plaque containing integer pi (1 ≤ pi ≤ n).

Little penguin Polo loves walking around this village. The walk looks like that. First he stands by a house number x

We know that:

1. When the penguin starts walking from any house indexed from 1 to k
   , inclusive, he can walk to house number 1.
2. When the penguin starts walking from any house indexed from k + 1 to n, inclusive, he definitely cannot walk to house number 1.
3. When the penguin starts walking from house number 1, he can get back to house number 1 after some non-zero number of walks from a house to a house.

You need to find the number of ways you may write the numbers on the houses' plaques so as to fulfill the three above described conditions. Print the remainder after dividing this number by 1000000007 (109 + 7).

Input

The single line contains two space-separated integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ min(8, n)) — the number of the houses and the number k from the statement.

Output

In a single line print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

5 2

Output

54

Input

7 4

Output

1728

題意：有n個點，給每個點x定義一個px值，表示從x點會到px點，現在有兩個要求：

1.從1到k的任意一點出發可以走到1點

2.從k+1到n的任意一點出發走不到1點

問滿足條件的p1,p2,...,pn個數

題解：分析到 k+1到n 門牌上肯定是 k+1到n的  前k個 就相當於是：1 是根節點 其餘點不斷往下放 dfs搜一遍即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int N=1e6+10;
typedef long long ll;
const int mod=1e9+7;
ll n,k;
ll C[10][10];
ll ksm(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=(ans*a)%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ans;
}
ll dfs(ll z,ll up)
{
    if(z<=0) return 1;
    ll ans=0;
    for(int i=1;i<=z;i++)
        ans=(ans+(C[z][i]*ksm(up,i)%mod)*dfs(z-i,i)%mod)%mod;
    return ans;
}
int main()
{
    C[0][0]=C[1][0]=C[1][1]=1;
    for(int i=2;i<=8;i++)
    {
        C[i][0]=1;
        for(int j=1;j<=8;j++)
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
    }
    while(~scanf("%lld%lld",&n,&k))
    {
        ll m=k-1;
        ll ans=k,cnt=0;
        cnt=dfs(k-1,1)%mod;
        ans=(ans*cnt)%mod;
        for(int i=1;i<=n-k;i++)
            ans=(ans*(n-k))%mod;
        cout<<ans<<endl;
    }
    return 0;
}