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 FOR(i, n, m) for(int i = n; i < (int)m; i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define pb push_back
using namespace std;
using ll = std::int_fast64_t;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 1000000007;
constexpr long double eps = 1e-15;
template<typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
    os << to_string(p.first) << " " << to_string(p.second);
    return os;
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
    REP(i, v.size()) {
        if(i) os << " ";
        os << to_string(v[i]);
    }
    return os;
}

struct modint {
    ll n;
public:
    modint(const ll n = 0) : n((n % mod + mod) % mod) {}
    static modint pow(modint a, int m) {
        modint r = 1;
        while(m > 0) {
            if(m & 1) { r *= a; }
            a = (a * a); m /= 2;
        }
        return r;
    }
    modint &operator++() { *this += 1; return *this; }
    modint &operator--() { *this -= 1; return *this; }
    modint operator++(int) { modint ret = *this; *this += 1; return ret; }
    modint operator--(int) { modint ret = *this; *this -= 1; return ret; }
    modint operator~() const { return (this -> pow(n, mod - 2)); } // inverse
    friend bool operator==(const modint& lhs, const modint& rhs) {
        return lhs.n == rhs.n;
    }
    friend bool operator<(const modint& lhs, const modint& rhs) {
        return lhs.n < rhs.n;
    }
    friend bool operator>(const modint& lhs, const modint& rhs) {
        return lhs.n > rhs.n;
    }
    friend modint &operator+=(modint& lhs, const modint& rhs) {
        lhs.n += rhs.n;
        if (lhs.n >= mod) lhs.n -= mod;
        return lhs;
    }
    friend modint &operator-=(modint& lhs, const modint& rhs) {
        lhs.n -= rhs.n;
        if (lhs.n < 0) lhs.n += mod;
        return lhs;
    }
    friend modint &operator*=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * rhs.n) % mod;
        return lhs;
    }
    friend modint &operator/=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * (~rhs).n) % mod;
        return lhs;
    }
    friend modint operator+(const modint& lhs, const modint& rhs) {
        return modint(lhs.n + rhs.n);
    }
    friend modint operator-(const modint& lhs, const modint& rhs) {
        return modint(lhs.n - rhs.n);
    }
    friend modint operator*(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * rhs.n);
    }
    friend modint operator/(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * (~rhs).n);
    }
};
istream& operator>>(istream& is, modint m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, modint m) { os << m.n; return os; }

/*
#define MAX_N 1000000
long long extgcd(long long a, long long b, long long& x, long long& y) {
    long long d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}
long long mod_inverse(long long a, long long m) {
    long long x, y;
    if(extgcd(a, m, x, y) == 1) return (m + x % m) % m;
    else return -1;
}
vector<long long> fact(MAX_N+1, inf);
long long mod_fact(long long n, long long& e) {
    if(fact[0] == inf) {
        fact[0]=1;
        if(MAX_N != 0) fact[1]=1;
        for(ll i = 2; i <= MAX_N; ++i) {
            fact[i] = (fact[i-1] * i) % mod;
        }
    }
    e = 0;
    if(n == 0) return 1;
    long long res = mod_fact(n / mod, e);
    e += n / mod;
    if((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
    return (res * fact[n % mod]) % mod;
}
// return nCk
long long mod_comb(long long n, long long k) {
    if(n < 0 || k < 0 || n < k) return 0;
    long long e1, e2, e3;
    long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2), a3 = mod_fact(n - k, e3);
    if(e1 > e2 + e3) return 0;
    return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}
 */

using mi = modint;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<mi> fact(1000, 0);
    fact[0] = 1;
    FOR(i, 1, 1000) fact[i] = fact[i - 1] * i;
    vector<mi> revfact(1000, 0);
    REP(i, 1000) revfact[i] = modint::pow(fact[i], mod - 2);
    vector<int> d(n);
    REP(i, n) cin >> d[i];
    vector<vector<vector<mi>>> dp(2, vector<vector<mi>>(n + 1, vector<mi>(2 * n + 1, 0)));
    // dp[i][j][k] =　頂点 i まで見て j 個閉路に加えて閉路外部への次数が k
    dp[0][0][0] = 1;
    REP(i, n) REP(j, n + 1) REP(k, 2 * n + 1) {
        dp[(i + 1) % 2][j][k] = 0;
        // 閉路に加える
        if(j && d[i] >= 2 && k - (d[i] - 2) >= 0) {
            dp[(i + 1) % 2][j][k] += dp[i % 2][j - 1][k - (d[i] - 2)] * revfact[d[i] - 2];
        }
        // 閉路に加えない
        dp[(i + 1) % 2][j][k] += dp[i % 2][j][k] * revfact[d[i] - 1];
    }
    mi ans = 0;
    FOR(i, 3, n) FOR(j, 1, 2 * n + 1) {
        ans += dp[n % 2][i][j] * fact[n - i - 1] * modint::pow(fact[j - 1], mod - 2) * fact[i - 1] * modint::pow(2, mod - 2) * fact[j];
    }
    ans += dp[n % 2][n][0] * fact[n - 1] * modint::pow(2, mod - 2);
    cout << ans << endl;
    return 0;
}
