Problem: 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

Solution (in CPP):

Solution:
#include <cstdio>
const int mo=1000000007;
int f[15][1<<15],dp[1<<15],n,m;
int calc(int x,int y){
	if (!x) return 1;
	long long cnt=1;
	for (int i=0;i<n && cnt;i++)
		if ((x>>i)&1) cnt=cnt*(f[i][y]-1)%mo;
			else if ((y>>i)&1) cnt=cnt*f[i][x]%mo;
	return cnt;
}
int main(){
	scanf("%d%d\n",&n,&m);
	int L=1<<n,ans=1;
	for (int i=0;i<n;i++)
		for (int j=0;j<L;j++) f[i][j]=1;
	for (int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d\n",&x,&y);
		x--;y--;(ans<<=1)%=mo;
		for (int j=0;j<L;j++)
			if ((j>>y)&1) f[x][j]<<=1;
	}
	dp[0]=1;
	for (int i=1;i<L;i++)
		for (int j=i;j;j=(j-1)&i)
			if (!((j>>1^j)&1))
				dp[i]=(dp[i]+1ll*calc(i^j,j)*dp[i^j])%mo;
	printf("%d\n",(ans-dp[L-1]+mo)%mo);
}