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>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long i64;
const int MOD = 1000000007;

int N, M, go[20];
i64 f[1 << 15], bin[300];

void update(i64 &x, i64 y) {
  x = (x + y) % MOD;
}

int main() {
  scanf("%d%d", &N, &M);
  
  for (int i = 1; i <= M; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    u = N - u, v = N - v;
    go[u] |= 1 << v;
  }
  
  bin[0]= 1;
  for (int i = 1; i <= M; ++i) bin[i] = (bin[i - 1] + bin[i - 1]) % MOD;
  
  int S = (1 << N) - 1;
  
  for (int mask = 1; mask <= S; ++mask) {
    int f1 = (mask >> N - 1) & 1, f2 = (mask >> N - 2) & 1;
    if (f1 ^ f2) continue;
    
    f[mask] = 1;
    for (int U = (mask - 1) & mask; U; U = (U - 1) & mask) {
      int T = mask ^ U;
      f1 = (T >> N - 1) & 1, f2 = (T >> N - 2) & 1;
      if (f1 ^ f2) continue;
      i64 way = 1;
      for (int i = 0; i < N; ++i)
	if ((T >> i) & 1) way = way * (bin[__builtin_popcount(go[i] & U)] - 1) % MOD;
	else if ((U >> i) & 1) way = way * bin[__builtin_popcount(go[i] & T)] % MOD;
      update(f[mask], f[T] * way % MOD);
    }
  }

  printf("%lld\n", (bin[M] - f[S] + MOD) % MOD);
  return 0;
}