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 N=20,M=5e4+50,mod=1e9+7;
int n,m,f[M],mx,b[N],c[N],pw[M],bit[M];
int main(){
	scanf("%d%d",&n,&m);mx=1<<n;f[0]=pw[0]=1;
	for(int i=1;i<=m;i++)pw[i]=pw[i-1]*2%mod;
	for(int i=1;i<mx;i++)bit[i]=bit[i>>1]+(i&1);
	for(int i=1,x,y;i<=m;i++){
		scanf("%d%d",&x,&y);x--;y--;
		b[x]|=1<<y;c[y]|=1<<x;
	}
	for(int i=1;i<mx;i++){
		if((i&1)^(i>>1&1))continue;
		if(i&1){
			for(int j=i;j;j=(j-1)&i){
				if((j&1)^(j>>1&1))continue;
				int dat=1,c2=0;
				for(int k=0;k<n;k++)if((i^j)>>k&1){
					int c1=bit[b[k]&j];c2+=bit[c[k]&j];
					dat=1ll*dat*(pw[c1]-1)%mod;
				}
				f[i]=(f[i]+1ll*f[i^j]*dat%mod*pw[c2])%mod;
			}
		}
		else{
			int c1=0;
			for(int j=0;j<n;j++)if(i>>j&1)c1+=bit[b[j]&i];
			f[i]=pw[c1];
		}
	}
	cout<<(pw[m]-f[mx-1]+mod)%mod;
	return 0;
}