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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define mmod 1000000007
#define N 16
#define M (1<<15)+10
using namespace std;
int n,m,a[N],f[M],bit[M],p2[N*N];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		a[x]|=1<<(y-1);
	}
	p2[0]=1;
	for(int i=1;i<=m;i++) p2[i]=p2[i-1]*2%mmod;
	int ss=(1<<n)-1;
	for(int i=1;i<=ss;i++) bit[i]=bit[i>>1]+(i&1);
	f[0]=1;
	for(int i=0;i<ss;i++)
	{
		if(f[i]==0) continue;
		int t=ss-i;
		for(int j=t;j;j=(j-1)&t)
			if((j&3)==0 || (j&3)==3)
			{
				int r=ss-i-j,cnt=0,tmp;
				for(int k=1;k<=n;k++) if(j&(1<<(k-1))) cnt+=bit[a[k]&r];
				tmp=p2[cnt];
				for(int k=1;k<=n;k++) if(r&(1<<(k-1))) tmp=1ll*tmp*(p2[bit[a[k]&j]]-1)%mmod;
				f[i+j]=(f[i+j]+1ll*f[i]*tmp)%mmod;
			}
	}
	int ans=(p2[m]-f[ss])%mmod;
	printf("%d\n",(ans+mmod)%mmod);
	return 0;
}