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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define MOD 1000000007
int n, m, f[1 << 15], g[15][1 << 15], h[15][1 << 15], bin[150];
inline void up(int &x, int y) {
	if ((x += y) >= MOD) x -= MOD;
}
inline int lowbit(int x) { return x & -x; }
int main() {
	scanf("%d%d", &n, &m);
	bin[0] = 1;
	for (int i = 1, u, v; i <= m; i++) {
		scanf("%d%d", &u, &v);
		u--, v--;
		g[u][1 << v] = 1;
		bin[i] = (bin[i - 1] << 1) % MOD;
	}
	for (int i = 0; i < n; i++) {
		for (int d = 1; d < (1 << n); d++) {
			g[i][d] = g[i][d - lowbit(d)] + g[i][lowbit(d)];
		}
	}
	f[0] = 1;
	for (int d = 1; d < (1 << n); d++) {
		for (int x = d; x; x = (x - 1) & d) {
			if ((x & 1) == ((x >> 1) & 1)) {
				int tmp = 1;
				for (int i = 0; i < n; i++) {
					if ((d >> i) & 1) {
						if ((x >> i) & 1) {
							tmp = 1LL * tmp * bin[g[i][d - x]] % MOD;
						} else {
							tmp = 1LL * tmp * (bin[g[i][x]] - 1) % MOD;
						}
					}
				}
				up(f[d], 1LL * tmp * f[d - x] % MOD);
			}
		}
	}
	printf("%d\n", (bin[m] - f[(1 << n) - 1] + MOD) % MOD);
	return 0;
}