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 <bits/stdc++.h>
using namespace std;

const int MOD = 1e9 + 7, N = 15;

namespace {
  int Add(int x, int y) { return (x += y) >= MOD ? x - MOD : x; }
  int Dec(int x, int y) { return (x -= y) < 0 ? x + MOD : x; }
  int Mul(int x, int y) { return 1LL * x * y % MOD; }
}

int n, m, Real;
int pw[N * N], f[1 << N], G[N];

int main(void) {
  scanf("%d%d", &n, &m);
  for(int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    x--; y--;
    G[x] |= 1 << y;
  }
  pw[0] = f[0] = 1;
  for(int i = 1; i <= m; i++)
    pw[i] = Mul(pw[i - 1], 2);		

  for(int S = 1; S < (1 << n); S++)
    if((S & 1) == (S >> 1 & 1))
      for(int T = S & (S - 1); ; T = S & (T - 1))
	if((T & 1) == (T >> 1 & 1)) {
	  int ways = 1;
	  for(int i = 0; i < n; i++)
	    if(S >> i & 1) {
	      if(T >> i & 1) ways = Mul(ways, Dec(pw[__builtin_popcount(G[i] & (S ^ T))], 1));
	      else ways = Mul(ways, pw[__builtin_popcount(G[i] & T)]);
	    }
	  f[S] = Add(f[S], Mul(ways, f[T]));
	  //cerr << S << " " << T << " " << ways << " " << f[S] << endl;
	  if(!T) break;
	}
  cout << Dec(pw[m], f[(1 << n) - 1]) << endl;
  return 0;
}
