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>
using namespace std;
int D[1 << 15];
int n, E[15], Mod = 1000000007, C[1 << 15], res = 1;
int main() {
	int i, j, m, a, b;
	scanf("%d%d", &n,&m);
	for (i = 0; i < m; i++) {
		scanf("%d%d", &a, &b);
		a--, b--;
		E[a] |= (1 << b);
		res = res * 2 % Mod;
	}
	for (i = 0; i < n; i++)for (j = 0; j < (1 << n); j++)if ((j >> i) & 1)C[j]++;
	D[0] = 1;
	for (i = 0; i < (1 << n); i++) {
		int bb = (1 << n) - 1 - i;
		if (!D[i])continue;
		for (int mask = bb; mask; mask = (mask - 1)&bb) {
			int rem = (1 << n) - 1 - i - mask;
			if ((mask & 3) && (mask & 3) != 3)continue;
			int r = D[i];
			for (j = 0; j < n; j++) {
				if (!((i | mask)&(1 << j))) {
					r = (long long)r * ((1 << C[E[j] & mask]) - 1) % Mod;
					continue;
				}
				if ((mask >> j) & 1) {
					r = ((long long)r << C[rem & E[j]]) % Mod;
				}
			}
			D[i | mask] = (D[i | mask] + r) % Mod;
		}
	}
	printf("%d\n", (res + Mod - D[(1 << n) - 1]) % Mod);
}