#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 2010;
const ll mod = 1000000007;
const ll inv2 = (mod + 1) / 2;

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

ll n, a[maxn], fac[maxn], ifac[maxn], f[55][maxn], g[maxn], pw[maxn], b[maxn];

inline ll C(ll n, ll m) {
	if (n < m || n < 0 || m < 0) {
		return 0;
	} else {
		return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
	}
}

void solve() {
	scanf("%lld", &n);
	fac[0] = pw[0] = 1;
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", &a[i]);
		fac[i] = fac[i - 1] * i % mod;
		pw[i] = pw[i - 1] * inv2 % mod;
		++b[a[i]];
	}
	ifac[n] = qpow(fac[n], mod - 2);
	for (int i = n - 1; ~i; --i) {
		ifac[i] = ifac[i + 1] * (i + 1) % mod;
	}
	for (int i = 0; i <= n; ++i) {
		for (int j = 0; j <= i; ++j) {
			g[i] = (g[i] + ((j & 1) ? mod - 1 : 1) * C(n, j) % mod * qpow(i - j + 1, n) % mod * ifac[n]) % mod;
		}
	}
	for (int i = n; i; --i) {
		g[i] = (g[i] - g[i - 1] + mod) % mod;
	}
	ll t = n - b[0];
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j <= t; ++j) {
			f[0][i + j] = (f[0][i + j] + g[i] * C(t, j) % mod * pw[t]) % mod;
		}
	}
	for (int i = 1; i <= 50; ++i) {
		int cnt = 0;
		for (int j = 1; j <= n; ++j) {
			cnt += (a[j] > i);
		}
		for (int j = 0; j <= n * 2; ++j) {
			if (!f[i - 1][j]) {
				continue;
			}
			for (int k = 0; k <= cnt; ++k) {
				int nj = (j >> 1) + k;
				f[i][nj] = (f[i][nj] + f[i - 1][j] * C(cnt, k) % mod * pw[cnt]) % mod;
			}
		}
	}
	ll ans = 0;
	for (int i = 0; i <= 50; ++i) {
		ll x = f[i][0];
		for (int j = i + 1; j <= 50; ++j) {
			int cnt = 0;
			for (int k = 1; k <= n; ++k) {
				cnt += (a[k] > j);
			}
			x = x * pw[cnt] % mod;
		}
		ans = (ans + b[i] * x) % mod;
	}
	ans = (mod + 1 - ans) % mod;
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}