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>
using namespace std;
const int mo=1e9+7;
int n,m,G[15],E[15],f[1<<15],bt[1<<15],S;
int main()
{
	int i,j,k,x,y;
	scanf("%d %d",&n,&m);
	for(i=1;i<=m;i++){
		scanf("%d %d",&x,&y);
		x--,y--,G[y]|=(1<<x);
		E[x]|=(1<<y);
	  }
	for(i=1;i<(1<<n);i++)
		bt[i]=bt[i-(i&(-i))]+1;
	f[0]=1,S=(1<<n)-1;
	for(i=1;i<(1<<n);i++)
		for(j=i;j;j=(j-1)&i){
			if((j&1)^((j>>1)&1)) continue;
			x=1;
			for(k=0;k<n;k++){
				if((j>>k)&1)
					x=(1LL*x*(1<<bt[G[k]&(i^j)]))%mo;
				if((~i>>k)&1)
					x=(1LL*x*((1<<bt[E[k]&j])-1))%mo;
			  }
			f[i]=(f[i]+1LL*f[i^j]*x)%mo;
		  }
	int ans=1;
	for(i=1;i<=m;i++) ans=(ans+ans)%mo;
	cout<<(ans+mo-f[(1<<n)-1])%mo;
	return 0;
}
