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>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int md=1000000007;
int n,m,i,j,k,g[1000],f[40000],num[40000],w[40000],u[40000],pw2[1000];
int main()
{
	scanf("%d%d",&n,&m);
	pw2[0]=1;
	for(i=1;i<=m;++i)
		pw2[i]=pw2[i-1]*2%md;
	for(i=1;i<(1<<n);++i)
		num[i]=num[i^(i&-i)]+1;
	for(i=1;i<=m;++i)
	{
		scanf("%d%d",&j,&k);
		g[j-1]|=1<<(k-1);
	}
	f[0]=1;
	for(i=1;i<(1<<n);++i)
	if((i&1)==(i>>1&1))
	{
		for(j=i;j>0;j=(j-1)&i)
		{
			w[j]=u[j]=1;
			for(k=0;k<n;++k)
				if(j>>k&1)
				{
					w[j]=1ll*w[j]*pw2[num[g[k]&(i^j)]]%md;
					u[j]=1ll*u[j]*(pw2[num[g[k]&(i^j)]]-1+md)%md;
				}
		}
		f[i]=1;
		for(j=(i-1)&i;j>0;j=(j-1)&i)
			f[i]=(f[i]+1ll*f[i^j]*w[j]%md*u[i^j]%md)%md;
	}
	printf("%d",(pw2[m]-f[(1<<n)-1]+md)%md);
	return 0;
}