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 Mod=1e9+7;

int N,M,Ans=1;
int To[15],F[1<<15];

int main(){
	int i,j,k,u,v,x;
	scanf("%d%d",&N,&M);
	while(M--){
		scanf("%d%d",&u,&v);
		To[u-1]|=1<<v-1;
		Ans=Ans*2%Mod;
	}
	F[0]=1;
	for(i=0;i<1<<N;i++)
		if((i&1)==(i>>1&1))
			for(j=i;j;j=j-1&i)
				if((j&1)==(j>>1&1)){
					for(x=1,k=0;k<N;k++)
						if(i>>k&1)
							x=1llu*x*((1<<__builtin_popcount(To[k]&(j>>k&1?i^j:j)))-(~j>>k&1))%Mod;
					F[i]=(F[i]+1llu*F[i^j]*x)%Mod;
				}
	printf("%d\n",(Ans-F[(1<<N)-1]+Mod)%Mod);
	return 0;
}