Problem: There is a directed graph G with N vertices and M edges. The vertices are numbered 1 through N, and the edges are numbered 1 through M. Edge i is directed from x_i to y_i. Here, x_i < y_i holds. Also, there are no multiple edges in G.

Consider selecting a subset of the set of the M edges in G, and removing these edges from G to obtain another graph G'. There are 2^M different possible graphs as G'.

Alice and Bob play against each other in the following game played on G'. First, place two pieces on vertices 1 and 2, one on each. Then, starting from Alice, Alice and Bob alternately perform the following operation:

* Select an edge i such that there is a piece placed on vertex x_i, and move the piece to vertex y_i (if there are two pieces on vertex x_i, only move one). The two pieces are allowed to be placed on the same vertex.



The player loses when he/she becomes unable to perform the operation. We assume that both players play optimally.

Among the 2^M different possible graphs as G', how many lead to Alice's victory? Find the count modulo 10^9+7.

Constraints

* 2 ≤ N ≤ 15
* 1 ≤ M ≤ N(N-1)/2
* 1 ≤ x_i < y_i ≤ N
* All (x_i,\ y_i) are distinct.

Input

Input is given from Standard Input in the following format:


N M
x_1 y_1
x_2 y_2
:
x_M y_M


Output

Print the number of G' that lead to Alice's victory, modulo 10^9+7.

Examples

Input

2 1
1 2


Output

1


Input

3 3
1 2
1 3
2 3


Output

6


Input

4 2
1 3
2 4


Output

2


Input

5 10
2 4
3 4
2 5
2 3
1 2
3 5
1 3
1 5
4 5
1 4


Output

816

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

using ll = long long;

const int MOD = 1e9 + 7;

int main() {
   ios_base::sync_with_stdio(false);
   int N, M;
   cin >> N >> M;
   vector<int> adj(N);
   for (int i = 0; i < M; ++i) {
      int v, u;
      cin >> v >> u;
      --v, --u;
      adj[v] |= (1 << u);
   }
   vector<int> popcnt(1 << N);
   for (int i = 1; i < (1 << N); ++i) {
      popcnt[i] = popcnt[i >> 1] + (i & 1);
   }
   vector<int> dp(1 << N);
   for (int mask = 1; mask < (1 << N); ++mask) {
      if ((mask & 3) == 1 || (mask & 3) == 2) continue;
      dp[mask] = 1;
      for (int posMask = (mask - 1) & mask; posMask > 0; posMask = (posMask - 1) & mask) {
         if ((posMask & 3) == 1 || (posMask & 3) == 2) continue;
         int zeroMask = mask ^ posMask;
         int ways = dp[posMask];
         for (int i = 0; i < N; ++i) {
            if (posMask >> i & 1) {
               ways = ll(ways) * ((1 << popcnt[adj[i] & zeroMask]) - 1) % MOD;
            } else if (zeroMask >> i & 1) {
               ways = ll(ways) * (1 << popcnt[adj[i] & posMask]) % MOD;
            }
         }
         dp[mask] += ways;
         if (dp[mask] >= MOD) dp[mask] -= MOD;
      }
   }
   int ans = 1;
   for (int i = 0; i < M; ++i) {
      ans += ans;
      if (ans >= MOD) ans -= MOD;
   }
   ans -= dp[(1 << N) - 1];
   if (ans < 0) ans += MOD;
   cout << ans << "\n";
   return 0;
}