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,mod=1e9+7;
int n,m,e[N],sz[1<<N],dp[1<<N];
int main()
{
	scanf("%d%d",&n,&m);int nn=1<<n;dp[0]=1;
	for(int i=1,u,v;i<=m;i++){scanf("%d%d",&u,&v);u--;v--;e[u]|=(1<<v);}
	for(int i=0;i<nn;i++)for(int j=0;j<n;j++)if(i>>j&1)sz[i]++;
	for(int i=0;i<nn;i++)sz[i]=1<<sz[i];
	for(int S=1;S<nn;S++)if((S&1)==(S>>1&1))for(int T=S;T;T=(T-1)&S)if((T&1)==(T>>1&1))
	{
		int r=1,U=S^T;
		for(int i=0;i<n;i++)if(T>>i&1)r=1ll*r*sz[e[i]&U]%mod;else if(U>>i&1)r=1ll*r*(sz[e[i]&T]-1)%mod;
		dp[S]=(dp[S]+1ll*dp[U]*r)%mod;
	}
	int ans=1;
	for(int i=1;i<=m;i++)ans=2*ans%mod;
	printf("%d\n",(ans+mod-dp[nn-1])%mod);
	return 0;
}