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 mo=1000000007;
const int N=32800;
int b[20],bit[N];
int f[N],n,m;
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		b[x]|=1<<(y-1);
	}
	for (int i=1;i<1<<n;i++)
		bit[i]=(i&1?bit[i>>1]*2+1:bit[i>>1]);
	f[0]=1;
	for (int i=1;i<1<<n;i++)
		if ((i&3)==0||(i&3)==3)
			for (int j=i;j;j=(j-1)&i){
				if ((j&3)!=0&&(j&3)!=3) continue;
				int tmp=1;
				for (int k=0;k<n;k++)
					if (i&(1<<k)){
						if (!(j&(1<<k)))
							tmp=1ll*tmp*bit[b[k+1]&j]%mo;
						else tmp=1ll*tmp*(bit[b[k+1]&(i^j)]+1)%mo;
					}
				f[i]=(f[i]+1ll*f[i^j]*tmp)%mo;
			}
	int ans=1;
	for (int i=1;i<=m;i++)
		ans=2*ans%mo;
	printf("%d",(ans+mo-f[(1<<n)-1])%mo);
}