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=1000000007;
void inline check(int &x){x-=p,x+=x>>31&p;}
int N,M;
int pow2[250];
int g[15][15];
int G[15][32768];

int F[32768];

int main(){
	scanf("%d%d",&N,&M);
	pow2[0]=1;for(int i=1;i<250;i++) check(pow2[i]=pow2[i-1]+pow2[i-1]);
	for(int i=1;i<=M;i++){
		int u,v;scanf("%d%d",&u,&v);u--;v--;
		g[u][v]=1;
	}
	for(int u=0;u<N;u++)
	for(int v=0;v<N;v++)
	for(int s=0;s<(1<<N);s++) if((s>>v)&1)
		G[u][s]+=g[u][v];
	for(int s=0;s<(1<<N);s++) if((s&3)==3){
		for(int A=s;A;A=(A-1)&s) if((A&3)==0){
			int B=s^A,tmp=F[B];
			for(int u=0;u<N;u++) if((B>>u)&1)
				tmp=1LL*tmp*(pow2[G[u][A]]-1)%p;
			for(int u=0;u<N;u++) if((A>>u)&1)
				tmp=1LL*tmp*pow2[G[u][B]]%p;
			check(F[s]+=tmp);
		}
		for(int A=s;A;A=(A-1)&s) if((A&3)==3){
			int B=s^A,tmp=1;
			for(int u=0;u<N;u++) if((B>>u)&1)
				tmp=1LL*tmp*(pow2[G[u][A]]-1)%p*pow2[G[u][B]]%p;
			for(int u=0;u<N;u++) if((A>>u)&1)
				tmp=1LL*tmp*pow2[G[u][B]]%p;
			check(F[s]+=tmp);
		}
	}
	printf("%d\n",(pow2[M]-F[(1<<N)-1]+p)%p);
}