Problem: We have N lamps numbered 1 to N, and N buttons numbered 1 to N. Initially, Lamp 1, 2, \cdots, A are on, and the other lamps are off.

Snuke and Ringo will play the following game.

* First, Ringo generates a permutation (p_1,p_2,\cdots,p_N) of (1,2,\cdots,N). The permutation is chosen from all N! possible permutations with equal probability, without being informed to Snuke.

* Then, Snuke does the following operation any number of times he likes:

* Choose a lamp that is on at the moment. (The operation cannot be done if there is no such lamp.) Let Lamp i be the chosen lamp. Press Button i, which switches the state of Lamp p_i. That is, Lamp p_i will be turned off if it is on, and vice versa.



At every moment, Snuke knows which lamps are on. Snuke wins if all the lamps are on, and he will surrender when it turns out that he cannot win. What is the probability of winning when Snuke plays optimally?

Let w be the probability of winning. Then, w \times N! will be an integer. Compute w \times N! modulo (10^9+7).

Constraints

* 2 \leq N \leq 10^7
* 1 \leq A \leq \min(N-1,5000)

Input

Input is given from Standard Input in the following format:


N A


Output

Print w \times N! modulo (10^9+7), where w is the probability of Snuke's winning.

Examples

Input

3 1


Output

2


Input

3 2


Output

3


Input

8 4


Output

16776


Input

9999999 4999


Output

90395416

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define SZ(x) ((int)x.size())
#define ALL(x) x.begin(),x.end()
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=(a); i<=(b); i++)
#define per(i,a,b) for (register int i=(a); i>=(b); i--)
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> Pii;
typedef vector<int> Vi;
template<class T> inline void read(T &x){
	x=0; char c=getchar(); int f=1;
	while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
	while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
template<class T> T gcd(T a, T b){return !b?a:gcd(b,a%b);}
template<class T> inline void umin(T &x, T y){x=x<y?x:y;}
template<class T> inline void umax(T &x, T y){x=x>y?x:y;}
inline ui R() {
	static ui seed=416;
	return seed^=seed>>5,seed^=seed<<17,seed^=seed>>13;
}
const int N = 1e7+11, mo = 1e9+7, inv2 =(mo+1)/2;
const ll MOD=7LL*mo*mo;
inline void add(int &x, int y){x=x+y<mo?x+y:x+y-mo;}
inline void sub(int &x, int y){x=x-y>=0?x-y:x-y+mo;}
inline int power(int a, int n) {
	int res=1;
	while (n) {
		if (n&1) res=1LL*res*a%mo;
		a=1LL*a*a%mo; n>>=1;
	}
	return res;
}

int fac[N],ifac[N];
void iniC(int n){
	fac[0]=1;rep(i,1,n)fac[i]=1LL*fac[i-1]*i%mo;
	ifac[n]=power(fac[n],mo-2);
	per(i,n-1,0)ifac[i]=1LL*ifac[i+1]*(i+1)%mo;
}
inline int C(int n, int m){return m>=0&&m<=n?1LL*fac[n]*ifac[m]%mo*ifac[n-m]%mo:0;}
int res;
int calc(int a, int t, int b){
	return 1ll*fac[a+b-1]*a%mo*fac[a+b+t]%mo*ifac[a+b]%mo;
}
int getans(int a, int t, int b){
	int res=0;rep(i,0,a)res=(res+1ll*(i&1?mo-1:1)*C(a,i)%mo*calc(a-i,t,b))%mo;
	return res;
}
int main() {
	int n,a;read(n);read(a);iniC(n+1);
	res=getans(a,0,n-a);
	rep(i,1,a)add(res,getans(i-1,a-i,n-a));
	cout<<res;
	return 0;
}