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>
using namespace std;
const int N=15;
const int mod=1e9+7;
typedef long long ll;
int n,m;
int f[1<<16];
int pw[(1<<15)+5];
int e[N];
int main()
{
	scanf("%d%d",&n,&m);
	pw[0]=1;f[0]=1;
	for(int i=1;i<=m;i++)pw[i]=(pw[i-1]<<1)%mod;
	for(int i=1;i<=m;i++){
		int u,v;scanf("%d%d",&u,&v);
		u--,v--;
		e[u]|=(1<<v);
	}
	for(int S=1;S<(1<<n);S++)if((S&1)==((S>>1)&1)){
		for(int T=S;T;T=(T-1)&S)if((T&1)==((T>>1)&1)){
			int ways=1;
			for(int i=0;i<n;i++)if((S>>i)&1){
				if((T>>i)&1)ways=1ll*ways*pw[__builtin_popcount(e[i]&(S^T))]%mod;
				else ways=1ll*ways*(pw[__builtin_popcount(e[i]&T)]-1)%mod;
			}
			f[S]=(f[S]+(ll)f[S^T]*ways)%mod;
			//cout<<T<<">"<<f[S^T]<<endl;
		}
		//cout<<S<<" "<<f[S]<<endl;
	}
	printf("%d\n",(pw[m]-f[(1<<n)-1]+mod)%mod);
}