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<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define NDEBUG
#define eprintf(...) do {} while (0)
#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}
template<unsigned MOD_> struct ModInt {
    static const unsigned MOD = MOD_;
    unsigned x;
    void undef() { x = (unsigned)-1; }
    bool isnan() const { return x == (unsigned)-1; }
    inline int geti() const { return (int)x; }
    ModInt() { x = 0; }
    ModInt(const ModInt &y) { x = y.x; }
    ModInt(int y) { if (y<0 || (int)MOD<=y) y %= (int)MOD; if (y<0) y += MOD; x=y; }
    ModInt(unsigned y) { if (MOD<=y) x = y % MOD; else x = y; }
    ModInt(long long y) { if (y<0 || MOD<=y) y %= MOD; if (y<0) y += MOD; x=y; }
    ModInt(unsigned long long y) { if (MOD<=y) x = y % MOD; else x = y; }
    ModInt &operator+=(const ModInt y) { if ((x += y.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(const ModInt y) { if ((x -= y.x) & (1u<<31)) x += MOD; return *this; }
    ModInt &operator*=(const ModInt y) { x = (unsigned long long)x * y.x % MOD; return *this; }
    ModInt &operator/=(const ModInt y) { x = (unsigned long long)x * y.inv().x % MOD; return *this; }
    ModInt operator-() const { return (x ? MOD-x: 0); }

    ModInt inv() const { return pow(MOD-2); }
    ModInt pow(long long y) const {
	ModInt b = *this, r = 1;
	if (y < 0) { b = b.inv(); y = -y; }
	for (; y; y>>=1) {
	    if (y&1) r *= b;
	    b *= b;
	}
	return r;
    }

    ModInt extgcd() const {
	unsigned a = MOD, b = x; int u = 0, v = 1;
	while (b) {
	    int t = a / b;
	    a -= t * b; swap(a, b);
	    u -= t * v; swap(u, v);
	}
	if (u < 0) u += MOD;
	return ModInt(u);
    }

    friend ModInt operator+(ModInt x, const ModInt y) { return x += y; }
    friend ModInt operator-(ModInt x, const ModInt y) { return x -= y; }
    friend ModInt operator*(ModInt x, const ModInt y) { return x *= y; }
    friend ModInt operator/(ModInt x, const ModInt y) { return x *= y.inv(); }
    friend bool operator<(const ModInt x, const ModInt y) { return x.x < y.x; }
    friend bool operator==(const ModInt x, const ModInt y) { return x.x == y.x; }
    friend bool operator!=(const ModInt x, const ModInt y) { return x.x != y.x; }
};

const LL MOD = 1000000007;
typedef ModInt<MOD> Mint;

const int MAX = 2011;
Mint inv[MAX], fact[MAX], fact_inv[MAX];

void init() {
    fact[0] = 1;
    for (int i=1; i<MAX; i++) fact[i] = fact[i-1] * i;
    fact_inv[MAX-1] = fact[MAX-1].inv();
    for (int i=MAX-2; i>=0; i--) fact_inv[i] = fact_inv[i+1] * (i+1);
    inv[0] = 0;
    for (int i=1; i<MAX; i++) inv[i] = fact_inv[i] * fact[i-1];
}

int N;
int D[311];
Mint buf[2][311][611];
Mint (*cur)[611] = buf[0];
Mint (*nxt)[611] = buf[1];

void MAIN() {
    init();
    scanf("%d", &N);
    REP (i, N) scanf("%d", D+i);

    if (*max_element(D, D+N) == 2) {
	Mint ans = fact[N-1] * inv[2];
	printf("%d\n", ans.geti());
	return;
    }
    
    cur[0][0] = 1;
    REP (i, N) cur[0][0] *= fact_inv[D[i]-1];
    
    REP (i, N) {
	memset(nxt, 0, sizeof buf[0]);
	
	REP (a, i+1) REP (b, N*2+1) if (cur[a][b].x) {
	    nxt[a][b] += cur[a][b];
	    if (D[i] >= 2) {
		nxt[a+1][b+D[i]-2] += cur[a][b] * fact[D[i]-1] * fact_inv[D[i]-2];
	    }
	}

	swap(cur, nxt);
    }
    
    Mint ans = 0;
    for (int a=3; a<=N; a++) for (int b=1; b<=2*N; b++) {
	    Mint tmp = cur[a][b] * fact[N-a-1] * fact_inv[b-1] * fact[a-1] * inv[2] * fact[b];
	    ans += tmp;
	}


    printf("%d\n", ans.geti());
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}

