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 <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 15, M = (1 << N) + 5, mo = 1e9 + 7 ;
using namespace std ;

bool G[N][N] ;
int n, m, fac[N * N + 5], A[N][M], B[N][M], f[M] ;

void upd(int &x, int y) {
	x = (x + y) % mo ;
}

int main() {
	scanf("%d%d", &n, &m) ;
	int x, y, st = (1 << n) - 1 ;
	fac[0] = 1 ;
	rep(i, 1, m) fac[i] = (ll) fac[i - 1] * 2 % mo ;
	rep(i, 1, m) {
		scanf("%d%d", &x, &y) ;
		G[x - 1][y - 1] = true ;
	}
	rep(i, 0, st) rep(j, 0, n - 1) if (!((i >> j) & 1)) {
		int cnt = 0 ;
		rep(k, 0, n - 1) if ((i >> k) & 1) cnt += G[j][k] ;
		B[j][i] = fac[cnt] - 1 ;
		A[j][i] = fac[cnt] ;
	}
	f[0] = 1 ;
	rep(i, 1, st) if (!(((i >> 0) & 1) ^ ((i >> 1) & 1))) {
		f[i] = 1 ;
		for (int j = i - 1 ; j > 0 ; j = (j - 1) & i) {
			int res = f[i ^ j] ;
			rep(k, 0, n - 1) if ((j >> k) & 1) res = (ll) res * A[k][i ^ j] % mo ;
			rep(k, 0, n - 1) if (!((j >> k) & 1) && ((i >> k) & 1)) res = (ll) res * B[k][j] % mo ;
			upd(f[i], res) ;
		}
	}
	int ans = (fac[m] - f[st]) % mo ;
	printf("%d\n", (ans + mo) % mo) ;
	return 0 ; 
}