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;
#define int long long
const int N=15,M=1e9+7;
int n,m,x,y,a[N][1<<N],f[1<<N],fac[N+1];
signed main(){
	scanf("%lld%lld",&n,&m);
	int sum=1;
	for (int i=1;i<=m;i++)(sum*=2)%=M;
	while (m--){
		scanf("%lld%lld",&x,&y);
		x--;y--;
		a[x][1<<y]++;
	}
	for (int i=0;i<n;i++)
		for (int j=0;j<n;j++)
			for (int k=0;k<1<<n;k++)
				if ((1<<j)&k)a[i][k]+=a[i][k^(1<<j)];
	fac[0]=1;
	for (int i=1;i<=n;i++)fac[i]=fac[i-1]*2;
	f[0]=1;
	for (int i=0;i<1<<n;i++)
		if ((i&3)==0||(i&3)==3)
			for (int j=i;j;j=(j-1)&i){
				int k=i^j,S=f[k];
				for (int l=0;l<n;l++)
					if ((1<<l)&j)(S*=fac[a[l][k]])%=M;
				for (int l=0;l<n;l++)
					if ((1<<l)&k)(S*=fac[a[l][j]]-1)%=M;
				(f[i]+=S)%=M;
			}
	printf("%lld\n",(sum+M-f[(1<<n)-1])%M);
}
