description: <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.
solution: 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());
          }
  
      }
  }
  
  