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>

typedef long long LL;
const int Mod = 1000000007;
const int MN = 15;

int N, M, A[MN][MN];
int w[MN * MN];
int c[1 << MN][MN], f[1 << MN];

int main() {
	scanf("%d%d", &N, &M), w[0] = 1;
	for (int i = 1; i <= M; ++i) w[i] = 2 * w[i - 1] % Mod;
	for (int i = 1, x, y; i <= M; ++i) scanf("%d%d", &x, &y), A[--x][--y] = 1;
	for (int S = 1; S < 1 << N; ++S) {
		int j = 0;
		while (~S >> j & 1) ++j;
		for (int u = 0; u < N; ++u)
			c[S][u] = c[S ^ 1 << j][u] + A[u][j];
	}
	for (int S = 0; S < 1 << N; ++S) if ((S & 3) == 3) {
		f[S] = 1;
		for (int T = S & (S - 1); T; --T &= S) if ((T & 1) == (T >> 1 & 1)) {
			if (T & 1) {
				int Coef = 1;
				for (int i = 0; i < N; ++i) if (S >> i & 1) {
					if (T >> i & 1) Coef = (LL)Coef * (w[c[S ^ T][i]] - 1) % Mod;
					else Coef = (LL)Coef * w[c[T][i]] % Mod;
				}
				f[S] = (f[S] + (LL)Coef * f[T]) % Mod;
			} else {
				int Coef = 1;
				for (int i = 0; i < N; ++i) if (S >> i & 1) {
					if (T >> i & 1) Coef = (LL)Coef * (w[c[S ^ T][i]] - 1) % Mod * w[c[T][i]] % Mod;
					else Coef = (LL)Coef * w[c[T][i]] % Mod;
				}
				f[S] = (f[S] + Coef) % Mod;
			}
		}
	}
	printf("%d\n", (w[M] - f[(1 << N) - 1] + Mod) % Mod);
	return 0;
}