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 <cstdio>

typedef long long i64;

const int N = 15, MOD = 1000000007;

int n, adj[N];
int ans, f[1 << N];

void init() {
  int m;
  scanf("%d%d", &n, &m);
  ans = 1;
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    --x, --y;
    adj[x] |= 1 << y;
    ans = 2 * ans % MOD;
  }
}

int main() {
  init();
  f[0] = 1;
  for (int s = 1; s < (1 << n); ++s) {
    if ((s ^ (s >> 1)) & 1) continue;
    for (int a = s; a; a = (a - 1) & s) {
      int b = s ^ a;
      int cur = f[b];
      if (!cur) continue;
      for (int i = 0; i < n; ++i) {
        if (a >> i & 1) cur = ((i64)cur << __builtin_popcount(adj[i] & b)) % MOD;
        if (b >> i & 1) cur = cur * ((1LL << __builtin_popcount(adj[i] & a)) - 1) % MOD;
      }
      f[s] = (f[s] + cur) % MOD;
    }
  }
  printf("%d\n", ((i64)ans - f[(1 << n) - 1] + MOD) % MOD);
  return 0;
}