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>
const int MOD = 1000000007;
int N, M, e[15][15], emask[15][32768];
long long f[8192];
long long A(int u, int v)
{
	long long W = 1;
	int zero = u - v, nonzero = v | 3 << N - 2;
	for (int i = 0; i < N; i++)
	{
		if (zero >> i & 1)
			W = (W << emask[i][nonzero]) % MOD;
		if (nonzero >> i & 1)
			W = W * ((1 << emask[i][zero]) - 1) % MOD;
	}
	return W;
}
long long B(int u, int v)
{
	long long W = 1;
	int zero = u - v | 3 << N - 2, nonzero = v;
	for (int i = 0; i < N; i++)
	{
		if (zero >> i & 1)
			W = (W << emask[i][nonzero]) % MOD;
		if (nonzero >> i & 1)
			W = (W << emask[i][nonzero]) % MOD * ((1 << emask[i][zero]) - 1) % MOD;
	}
	return W;
}
int main()
{
	scanf("%d%d", &N, &M);
	long long BA = 1;
	for (int i = 1, u, v; i <= M; i++)
	{
		scanf("%d%d", &u, &v);
		e[N - u][N - v] = 1;
		BA = (BA << 1) % MOD;
	}
	for (int i = 0; i < N; i++)
		for (int j = 0; j < 1 << N; j++)
			for (int k = 0; k < N; k++)
				if ((j >> k & 1) && e[i][k])
					emask[i][j]++;
	for (int i = 0; i < 1 << N - 2; i++)
	{
		f[i] = B(i, i);
		for (int j = 0; j < i; j = ((j | ~i) + 1) & i)
			f[i] = (f[i] + f[j] * A(i, j) + B(i, j)) % MOD;
	}
	printf("%lld\n", (BA - f[(1 << N - 2) - 1] + MOD) % MOD);
	return 0;
}