1556_F. Sports Betting_170
<image>

William is not only interested in trading but also in betting on sports matches. n teams participate in each match. Each team is characterized by strength a_i. Each two teams i < j play with each other exactly once. Team i wins with probability (a_i)/(a_i + a_j) and team j wins with probability (a_j)/(a_i + a_j).

The team is called a winner if it directly or indirectly defeated all other teams. Team a defeated (directly or indirectly) team b if there is a sequence of teams c_1, c_2, ... c_k such that c_1 = a, c_k = b and team c_i defeated team c_{i + 1} for all i from 1 to k - 1. Note that it is possible that team a defeated team b and in the same time team b defeated team a.

William wants you to find the expected value of the number of winners.

Input

The first line contains a single integer n (1 ≤ n ≤ 14), which is the total number of teams participating in a match.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — the strengths of teams participating in a match.

Output

Output a single integer — the expected value of the number of winners of the tournament modulo 10^9 + 7.

Formally, let M = 10^9+7. It can be demonstrated that the answer can be presented as a irreducible fraction p/q, where p and q are integers and q not ≡ 0 \pmod{M}. Output a single integer equal to p ⋅ q^{-1} mod M. In other words, output an integer x such that 0 ≤ x < M and x ⋅ q ≡ p \pmod{M}.

Examples

Input


2
1 2


Output


1


Input


5
1 5 2 11 14


Output


642377629

Note

To better understand in which situation several winners are possible let's examine the second test:

One possible result of the tournament is as follows (a → b means that a defeated b):

  * 1 → 2 
  * 2 → 3 
  * 3 → 1 
  * 1 → 4 
  * 1 → 5 
  * 2 → 4 
  * 2 → 5 
  * 3 → 4 
  * 3 → 5 
  * 4 → 5 



Or more clearly in the picture:

<image>

In this case every team from the set \{ 1, 2, 3 \} directly or indirectly defeated everyone. I.e.:

  * 1st defeated everyone because they can get to everyone else in the following way 1 → 2, 1 → 2 → 3, 1 → 4, 1 → 5. 
  * 2nd defeated everyone because they can get to everyone else in the following way 2 → 3, 2 → 3 → 1, 2 → 4, 2 → 5. 
  * 3rd defeated everyone because they can get to everyone else in the following way 3 → 1, 3 → 1 → 2, 3 → 4, 3 → 5. 



Therefore the total number of winners is 3.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskF solver = new TaskF();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF {
        final int MOD = (int) (1e9 + 7);

        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
            }
            int[][] p = new int[n][n];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    p[i][j] = (int) (a[i] * (long) inv(a[i] + a[j]) % MOD);
                }
            }

            long[] probIsScc = new long[1 << n];
            long ans = 0;
            for (int sccMask = 0; sccMask < 1 << n; sccMask++) {
                long probLeadingScc = 1;
                for (int i = 0; i < n; i++) {
                    if (((sccMask >> i) & 1) == 0) {
                        continue;
                    }
                    for (int j = 0; j < n; j++) {
                        if (((sccMask >> j) & 1) != 0) {
                            continue;
                        }
                        probLeadingScc = probLeadingScc * p[i][j] % MOD;
                    }
                }

                probIsScc[sccMask] = 1;
                for (int m = (sccMask - 1) & sccMask; m > 0; m = (m - 1) & sccMask) {
                    long x = probIsScc[m];
                    for (int i = 0; i < n; i++) {
                        if (((m >> i) & 1) == 0) {
                            continue;
                        }
                        for (int j = 0; j < n; j++) {
                            if (((sccMask >> j) & 1) == 0) {
                                continue;
                            }
                            if (((m >> j) & 1) != 0) {
                                continue;
                            }
                            x = x * p[i][j] % MOD;
                        }
                    }

                    probIsScc[sccMask] -= x;
                    if (probIsScc[sccMask] < 0) {
                        probIsScc[sccMask] += MOD;
                    }
                }

                long cur = 1;
                cur = cur * probIsScc[sccMask] % MOD;
                cur = cur * probLeadingScc % MOD;
                cur = cur * Integer.bitCount(sccMask) % MOD;
                ans = (ans + cur) % MOD;
            }
            out.println(ans);
        }

        private int inv(int x) {
            return pow(x, MOD - 2);
        }

        private int pow(long a, long n) {
            long r = 1;
            while (n > 0) {
                if (n % 2 == 1) {
                    r = r * a % MOD;
                }
                a = a * a % MOD;
                n /= 2;
            }
            return (int) r;
        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}

