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<bits/stdc++.h>
using namespace std;
const int Max_N=15;
const long long Mod=(1e9)+7;
int n,m,a[Max_N+10][Max_N+10],P[1<<Max_N][Max_N+5];
long long f[1<<Max_N];
inline long long Mul(int x,int y){
	long long a=1,b=x;
	for(;y;y>>=1,b=b*b%Mod) if(y&1) a=a*b%Mod;
	return a;
}
int main(){
	scanf("%d%d",&n,&m);
	int all=(1<<n)-1;
	for(int i=1;i<=m;i++){
		int x,y; scanf("%d%d",&x,&y);
		a[x][y]=1;
	}
	for(int i=0;i<=all;i++)
	 for(int j=1;j<=n;j++)
	  if(!((1<<j-1)&i))
	   for(int k=1;k<=n;k++)
		if(((1<<k-1)&i)&&a[j][k])
		 ++P[i][j];
	for(int i=1;i<=all;i++){
		long long tot=1;
		for(int j=(i-1)&i;j;j=(j-1)&i){
		 	if((1&j)&&(2&(i^j))) continue;
		 	if((2&j)&&(1&(i^j))) continue;
			long long sum=1; int e=0;
		 	for(int k=1;k<=n;k++)
		 	 if((1<<k-1)&(i^j))
		 	  sum=sum*((Mul(2,P[j][k])-1+Mod)%Mod)%Mod;
		 	for(int k=1;k<=n;k++)
		 	 if((1<<k-1)&j)
		 	  e+=P[i^j][k]; 
		 	tot+=sum*f[i^j]%Mod*Mul(2,e)%Mod,tot%=Mod;
		 }
		f[i]=tot;
	}
	printf("%lld\n",(Mul(2,m)-f[(1<<n)-1]+Mod)%Mod);
	return 0;
}