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<bits/stdc++.h>
#define mo 1000000007
using namespace std;
int bc[1<<15],f[1<<15],des[15];
int main()
{
	int n,m,d=1;scanf("%d%d",&n,&m);
	while (m--)
	{
		int u,v;scanf("%d%d",&u,&v);u--;v--;
		des[u]|=(1<<v);d=d*2%mo;
	}
	for (int i=1;i<(1<<n);i++) bc[i]=bc[i>>1]+(i&1);
	f[0]=1;
	for (int s=1;s<(1<<n);s++) if ((s&1)==((s>>1)&1))
	for (int t=s;t;t=(t-1)&s) if ((t&1)==((t>>1)&1))
	{
		int js=1;
		for (int i=0;i<n;i++) if ((1<<i)&s)
		{
			if ((1<<i)&t) js=(long long)js*(1<<bc[des[i]&(s-t)])%mo;
			else js=(long long)js*((1<<bc[des[i]&t])-1)%mo;
		}
		f[s]=(f[s]+(long long)js*f[s-t])%mo;
	}
	printf("%d\n",(d+mo-f[(1<<n)-1])%mo);
}