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,M,f[1<<15],g[15][2];
bool G[15][15];
int dfs(int s,int i){
	while(i>=0&&!(s>>i&1))--i;
	if(i<0)return f[s];
	for(int j=0;j<i;j++)if(G[j][i])++g[j][0];
	long long res=dfs(s,i-1);
	res=(res<<g[i][1])-res;
	for(int j=0;j<i;j++)if(G[j][i])--g[j][0],++g[j][1];
	res+=1ll*dfs(s^1<<i,i-1)<<g[i][0];
	for(int j=0;j<i;j++)if(G[j][i])--g[j][1];
	return res%mod;
}
int main(){
	scanf("%d%d",&N,&M);
	for(int i=0,a,b;i<M;i++)scanf("%d%d",&a,&b),G[--a][--b]=1;
	f[0]=1;
	for(int i=0;i<N;i++){
		for(int s=0;s<1<<i;s++)if(((s|1<<i)&1)==((s|1<<i)>>1&1)){
			for(int j=0;j<i;j++)g[j][0]=0,g[j][1]=G[j][i];
			f[s|1<<i]=dfs(s,i);
		}
	}
	int s=1;
	while(M--)s=s*2%mod;
	s=(s+mod-f[(1<<N)-1])%mod;
	printf("%d\n",s);
}
