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
#include <cstdio>
#include <iostream>
#define ri register int
using namespace std;

const int N = 15;
const int M = 1 << N | 3;
const int mod = 1e9 + 7;

int n, m;
int G[N], f[M], cnt[M];
int tab[M][N], top[M];

inline void prt(int x) {
  if (!x) return;
  prt(x >> 1);
  cerr << (x & 1);
}

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, x, y; i <= m; i ++) {
    scanf("%d%d", &x, &y);
    G[x - 1] |= 1 << (y - 1);
  }
  int tot = 1;
  for (int i = 1; i <= m; i ++) tot = (tot << 1) % mod;
  for (int i = 1; i < (1 << N); i ++)
    cnt[i] = cnt[i - (i & (- i))] + 1;
  int U = 1 << n;
  for (int i = 1; i < U; i ++)
    for (int j = 0; j < n; j ++)
      if ((i >> j) & 1) tab[i][++ top[i]] = j;
  for (int i = 0; i < U; i ++)
    if ((i & 1) == ((i >> 1) & 1)) {
      for (int j = i; j; j = (j - 1) & i)
        if (f[j]) {
          ri w = i ^ j, res = 1;
          for (ri k = top[j]; k && res; k --)
            res = ((1LL << cnt[G[tab[j][k]] & w]) - 1) * res % mod;
          for (ri k = top[w]; k && res; k --)
            res = (1LL << cnt[G[tab[w][k]] & j]) * res % mod;
          f[i] = (f[i] + 1LL * f[j] * res) % mod;
        }
      f[i] ++;
    }
  printf("%d\n", (tot - f[U - 1] + mod) % mod);
  return 0;
}
