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 P=1e9+7;

int n,m,k,go[15],bcnt[1<<15],dp[1<<15],pw[233];
bool fb[1<<15];

int main(){
	scanf("%d%d",&n,&m);
	k=1<<n,pw[0]=1;
	for(int i=1,u,v;i<=m;i++)scanf("%d%d",&u,&v),u--,v--,go[u]|=1<<v,pw[i]=(pw[i-1]<<1)%P;
	for(int i=1;i<k;i++)bcnt[i]=bcnt[i>>1]+(i&1),fb[i]=(i&3)==2||(i&3)==1;
	for(int S=1;S<k;S++)if(!fb[S]){
		int&w=dp[S];w=1;
		for(int T=S;T;T=(T-1)&S)if(!fb[T]){
			int V=S-T;
			int a=1;
			for(int i=0;i<n;i++){
				if(V>>i&1)a=(long long)a*(pw[bcnt[go[i]&T]]-1)%P;
				if(T>>i&1)a=(long long)a*pw[bcnt[go[i]&V]]%P;
			}
			w=(w+(long long)a*dp[V])%P;
		}
	}
	printf("%lld\n",(pw[m]-dp[k-1]+P)%P);
	return 0;
}