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<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 20
#define mo 1000000007
#define ll long long
using namespace std;
int n,m,a[N],len[32768],bz[32768];
ll f[32768],e[N*N];
int main()
{
	scanf("%d%d",&n,&m);
	e[0]=1;fo(i,1,15*7) e[i]=e[i-1]*2%mo;
	fo(i,1,m)
	{
		int x,y;scanf("%d%d",&x,&y);
		a[x]|=e[y-1];
	}
	fo(i,0,e[n]-1)
	{
		fo(j,1,n) if((i&e[j-1])>0) len[i]++;
		bz[i]=(((i&1)>0)==((i&2)>0));
	}
	f[0]=1;
	fo(s,1,e[n]-1)
	if(bz[s])
	{
		for(int t=s;;t=(t-1)&s)
		if(bz[t])
		{
			ll jy=1;
			fo(i,1,n)
			{
				if(((s-t)&e[i-1])>0) jy=(jy*(e[len[a[i]&t]]-1))%mo;
				if((t&e[i-1])>0) jy=jy*e[len[a[i]&(s-t)]]%mo;
			}
			f[s]=(f[s]+jy*f[s-t])%mo;
			if(t==0) break;
		}
	}
	ll ans=e[m];
	ans=(ans-f[e[n]-1]+mo)%mo;
	printf("%lld\n",ans);
}