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>
typedef long long i64;
const int P=1e9+7;
int n,m,pw2[337],mx,e1,e2,f[8207][3],c1[8207],e[17];
int main(){
	scanf("%d%d",&n,&m);
	n-=2;
	mx=1<<n;
	pw2[0]=1;
	for(int i=1;i<mx;++i)c1[i]=c1[i>>1]+(i&1);
	for(int i=1;i<=300;++i)pw2[i]=pw2[i-1]*2%P;
	for(int i=1,a,b;i<=m;++i){
		scanf("%d%d",&a,&b);
		if(a!=1||b!=2)(a==1?e1:a==2?e2:e[a-3])|=1<<b-3;
	}
	f[0][0]=f[0][2]=1;
	for(int i=1,mi=1;i<mx;++i){
		if(!(mi&i))mi<<=1;
		for(int s0=i;s0;s0=s0-1&i)if(s0&mi){
			int s1=i^s0;
			i64 e10=1;
			int ce=0;
			for(int j=0;j<n;++j)if(s1>>j&1)e10=e10*(pw2[c1[s0&e[j]]]-1)%P;
			else if(s0>>j&1)ce+=c1[s1&e[j]];
			f[i][0]=(f[i][0]+e10*pw2[ce]%P*f[s1][0])%P;
			f[i][2]=(f[i][2]+e10*pw2[ce+c1[s1&e1]+c1[s1&e2]]%P*f[s1][0])%P;
			e10=e10*(pw2[c1[s0&e1]]-1)%P*(pw2[c1[s0&e2]]-1)%P;
			f[i][1]=(f[i][1]+e10*pw2[ce]%P*(f[s1][2]+f[s1][1]))%P;
		}
	}
	printf("%d\n",((pw2[m]-(f[mx-1][2]+f[mx-1][1])%P)%P+P)%P);
	return 0;
}