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<cstdlib>
#include<cstring>
#include<algorithm>
#define fo(i,j,k) for(int i=j;i<=k;++i)
#define fd(i,j,k) for(int i=j;i>=k;--i)
#define ll long long
using namespace std;
const int N=16,M=300,mo=1e9+7;
ll f[1<<N];
int mp[N][1<<N];
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	fo(i,1,m){
		int x,y;
		scanf("%d %d",&x,&y);
		mp[x][1<<y-1]=1;
	}
	fo(i,1,n)
	fo(j,1,(1<<n)-1) mp[i][j]=mp[i][j-(j&-j)]+mp[i][(j&-j)];
	f[0]=1;
	fo(S,1,(1<<n)-1) if((S&1)==((S&2)>>1)){
		f[S]=1;
		for(int T=(S-1)&S;T;T=(T-1)&S) if((T&1)==((T&2)>>1)){
			int x=S^T;
			ll t=f[x];
			if(!t) continue;
			fo(i,1,n) if(x&(1<<i-1)) t=t*((1<<mp[i][T])-1)%mo;
			else if(T&(1<<i-1)) t=t*(1<<mp[i][x])%mo;
			f[S]=(f[S]+t)%mo;
		}
	}
	ll ans=1;
	fo(i,1,m) ans=ans*2%mo;
	printf("%lld",(ans-f[(1<<n)-1]+mo)%mo);
}