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>
const int Mod = 1e9 + 7;
int f[1 << 15], n, m, u[110], v[110], P[110], s[15], d[1 << 15], to[1 << 15];
int main()
{
	scanf("%d %d", &n, &m);
	P[0] = 1; for(int i = 1; i <= m; ++i) P[i] = (P[i - 1] << 1) % Mod;
	for(int i = 1; i <= m; ++i) scanf("%d %d", &u[i], &v[i]);
	int Full = 1 << n;
	for(int i = 1; i < Full; ++i) d[i] = d[i - (-i & i)] + 1;
	for(int i = 1; i <= m; ++i) to[u[i] - 1] |= 1 << v[i] - 1;
	for(int i = 0; i < Full; ++i)
	{
		if(i & 1 ^ i >> 1 & 1);
		else f[i] = 1;
		int t = (i - 1) & i;
		while(t)
		{
			if(i - t & 1 ^ i - t >> 1 & 1 || !f[i - t]);
			else
			{
				int c1 = 0, c2 = 0, tmp = 0;
				for(int j = 0; j < n; ++j) if(i - t >> j & 1) s[j] = 0;
				for(int j = 0; j < n; ++j) 
					if(i - t >> j & 1) s[j] += d[to[j] & t];
					else if(t >> j & 1) c2 += d[to[j] & i - t];
				int g = 1ll * P[c2] * f[i - t] % Mod;
				for(int j = 0; j < n; ++j) if(i - t >> j & 1)
				{
					if(s[j]){ g = 1ll * g * (P[s[j]] - 1) % Mod; }
					else { g = 0; break; }
				}
				f[i] = (f[i] + g) % Mod;
			}
			t = (t - 1) & i;
		}
	}
	printf("%d\n", (P[m] - f[Full - 1] + Mod) % Mod);
	return 0;
}