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>
const int mod=1000000007;
int N,deg[20],g[20],cnt[2][20];
bool G[20][20];
int dfs(int i){
	if(i==1){
		int t=deg[0]+deg[1],s=1<<t,w=1;
		if(G[0][1])--t;
		for(i=0;i<N&&w;i++){
			t-=cnt[0][i]+cnt[1][i];
			s-=w<<t;
			w=(w<<cnt[0][i])-w;
			w=(w<<cnt[1][i])-w;
		}
		return s;
	}
	int t=deg[i],c[t+1];
	long long s=0,w=1;
	for(int j=0;j<=deg[i];j++)c[j]=0;
	for(int j=i+1;j<N;j++)if(G[i][j]&&g[j]<=deg[i])c[g[j]]++;
	for(int j=0;j<=deg[i]&&w;j++){
		t-=c[j];
		g[i]=j;
		if(G[0][i])cnt[0][j]++;
		if(G[1][i])cnt[1][j]++;
		(s+=w*dfs(i-1)%mod<<t)%=mod;
		w=((w<<c[j])-w)%mod;
		if(G[0][i])cnt[0][j]--;
		if(G[1][i])cnt[1][j]--;
	}
	return s;
}
int main(){
	int M;scanf("%d%d",&N,&M);
	while(M--){
		int x,y;scanf("%d%d",&x,&y);
		G[--x][--y]=1;
		deg[x]++;
	}
	printf("%d\n",dfs(N-1));
}