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<iostream>
#include<cstdio>
#include<cstring>
#define ll long long
#define mod 1000000007
#define N 510
using namespace std;
ll f[N][N][2],fac[N],inv[N];
ll ksm(ll a,ll b=mod-2)
{
    ll r=1;
    for(;b;b>>=1)
    {
        if(b&1) r=r*a%mod;
        a=a*a%mod;
    }
    return r;
}
int d[N];
int main()
{
    int n;
    scanf("%d",&n);
    if(n<=2){puts("0");return 0;}
    fac[0]=1;
    for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
    inv[n]=ksm(fac[n]);
    for(int i=n-1;i>=0;i--) inv[i]=inv[i+1]*(i+1)%mod;
    for(int i=1;i<=n;i++) scanf("%d",&d[i]);
    f[0][0][0]=1;
    for(int i=1;i<=n;i++)
        for(int j=0;j<=i;j++)
        {
            (f[i][j][0]+=f[i-1][j][0]*inv[d[i]-1])%=mod,(f[i][j][1]+=f[i-1][j][1]*inv[d[i]-1])%=mod;
            if(d[i]>=2 && j) (f[i][j][0]+=f[i-1][j-1][0]*inv[d[i]-2])%=mod,(f[i][j][1]+=f[i-1][j-1][1]*inv[d[i]-2])%=mod;
            if(d[i]>=3 && j) (f[i][j][1]+=f[i-1][j-1][0]*inv[d[i]-3])%=mod;
        }
    ll ans=0;
	for(int i=3;i<n;i++)
		ans=(ans+f[n][i][1]*fac[n-i-1]%mod*fac[i-1]%mod*inv[2]%mod)%mod;
    printf("%lld\n",(ans+f[n][n][0]*fac[n-1]%mod*inv[2])%mod);
    return 0;
}