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 L long long
#define pb push_back
#define pi pair<int,int>
#define pii pair<pair<int,int>,int>
#define aa first
#define bb second
#define xx aa.aa
#define yy aa.bb
#define zz bb
#define mp make_pair
#define mpp(a,b,c) mp(mp(a,b),c)
#define g(a,b) g[h[a]+h[(1<<n)-1^b]]
using namespace std;
const int q=1000000007;
int n,m,x[20][20],f[40000],g[15000000],h[40000];
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k,l,u,w;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++)
	  {
	   scanf("%d%d",&j,&k);
	   j--;
	   k--;
	   x[j][k]=1;
	  }
	for(i=0;i<(1<<n);i++)
	  h[i]=h[i>>1]*3+(i&1);
	for(i=1;i<(1<<n);i++)
	  for(j=(i-1&i);j;j=(j-1&i))
	    {
		 k=i-j;
		 for(l=0;!(k&(1<<l));l++);
		 if(k==(1<<l))
		   {
		    for(w=0,u=0;u<n;u++)
		      if(j&(1<<u) && x[l][u])
		        w=(w*2+1)%q;
		    for(u=0;u<n;u++)
		      if(j&(1<<u) && x[u][l])
		        w=(w<<1)%q;
		   }
		 else
		   w=(L)g(j,1<<l)*g(j,k^(1<<l))%q;
		 g(j,k)=w;
		}
	for(i=0;i<(1<<n);i++)
	  if(!((i&1)^((i>>1)&1)))
	    {
		 f[i]=1;
		 for(j=i-1&i;j;j=(j-1&i))
		   f[i]=(f[i]+(L)f[j]*g(i^j,j))%q;
		}
	for(i=1,k=1;i<=m;i++)
	  k=(k<<1)%q;
	printf("%d\n",(k-f[(1<<n)-1]+q)%q);
	return 0;
}
