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 <bits/stdc++.h>
using namespace std;

typedef long long  LL;
const int mo=int(1e9+7);
int n,m,a[16],b[16],c[1<<15],f[1<<15],ans;

void work()
{
	scanf("%d %d",&n,&m);
	for (int i=1,x,y; i<=m; i++)
		scanf("%d %d",&x,&y),a[x]|=1<<(y-1),b[y]|=1<<(x-1);
	for (int i=1; i<(1<<n); i++)  c[i]=c[i/2]+(i&1);
	f[0]=1;
	for (int i=1; i<(1<<n); i++)
		for (int j=i; j; j=(j-1)&i)
			if (c[j&3]!=1)
				{
					LL s=1;
					for (int k=1; k<=n; k++)
						{
							if (j&(1<<(k-1)))  (s<<=c[b[k]&(i^j)])%=mo;
							if (!(i&(1<<(k-1))))  s=s*((1<<c[a[k]&j])-1)%mo;
						}
					f[i]=(f[i]+f[i^j]*s)%mo;
				}
	ans=1;
	for (int i=1; i<=m; i++)  (ans<<=1)%=mo;
	printf("%d",(ans+mo-f[(1<<n)-1])%mo);
}

int main()
{
	work();
	return 0;
}