Snuke has come up with the following problem.

> You are given a sequence d of length N. Find the number of the undirected graphs with N vertices labeled 1,2,...,N satisfying the following conditions, modulo 10^{9} + 7:
>
>   * The graph is simple and connected.
>   * The degree of Vertex i is d_i.
>


When 2 \leq N, 1 \leq d_i \leq N-1, {\rm Σ} d_i = 2(N-1), it can be proved that the answer to the problem is \frac{(N-2)!}{(d_{1} -1)!(d_{2} - 1)! ... (d_{N}-1)!}.

Snuke is wondering what the answer is when 3 \leq N, 1 \leq d_i \leq N-1, { \rm Σ} d_i = 2N. Solve the problem under this condition for him.

Constraints

* 3 \leq N \leq 300
* 1 \leq d_i \leq N-1
* { \rm Σ} d_i = 2N

Input

Input is given from Standard Input in the following format:


N
d_1 d_2 ... d_{N}


Output

Print the answer.

Examples

Input

5
1 2 2 3 2


Output

6


Input

16
2 1 3 1 2 1 4 1 1 2 1 1 3 2 4 3


Output

555275958
#include<bits/stdc++.h>
#define mod 1000000007
using namespace std;
int n,d[305];
int dp[305][305][2],ans;
int fac[305],ifac[305];

int ksm(int x,int y){
	int res=1;
	while(y){
		if(y&1) res=1ll*res*x%mod;
		x=1ll*x*x%mod,y/=2;
	}
	return res;
}

int main(){
	scanf("%d",&n);
	int flag=1;
	for(int i=1;i<=n;i++) scanf("%d",&d[i]),flag&=(d[i]==2);
	dp[0][0][0]=1;
	fac[0]=1;for(int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mod;
	ifac[n]=ksm(fac[n],mod-2);
	for(int i=n-1;i>=0;i--) ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
	if(flag){
		if(n>=3) printf("%d\n",1ll*fac[n-1]*ifac[2]%mod);
		else puts("0");return 0;
	}
	for(int i=1;i<=n;i++){
		for(int j=0;j<i;j++){
			if(d[i]>=3) (dp[i][j+1][1]+=1ll*ifac[d[i]-3]*dp[i-1][j][0]%mod)%=mod;
			if(d[i]>=2){
				(dp[i][j+1][0]+=1ll*ifac[d[i]-2]*dp[i-1][j][0]%mod)%=mod;
				(dp[i][j+1][1]+=1ll*ifac[d[i]-2]*dp[i-1][j][1]%mod)%=mod;
			}
			(dp[i][j][0]+=1ll*ifac[d[i]-1]*dp[i-1][j][0]%mod)%=mod;
			(dp[i][j][1]+=1ll*ifac[d[i]-1]*dp[i-1][j][1]%mod)%=mod;
		}
	}
	for(int i=3;i<n;i++) (ans+=1ll*dp[n][i][1]*fac[i-1]%mod*ifac[2]%mod*fac[n-i-1]%mod)%=mod;
	printf("%d\n",ans);

	return 0;
}