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<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=beg[j];i;i=nex[i])
using namespace std;
int const mn=15+3,ms=(1<<15)+3,mo=1e9+7;
int n,m,f[ms],g[mn][ms],pw2[mn*mn];
int main(){
	scanf("%d%d",&n,&m);
	fo(i,1,m){
		int u,v;
		scanf("%d%d",&u,&v);
		g[u][1<<(v-1)]=1;
	}
	pw2[0]=1;
	fo(i,1,n*(n-1)/2)pw2[i]=2ll*pw2[i-1]%mo;
	int lim=(1<<n)-1;
	fo(i,1,n)fo(s,1,lim)for(int ss=(s-1)&s;ss;ss=(ss-1)&s)
		g[i][s]=g[i][ss]+g[i][s-ss];
	fo(s,0,lim)if((!(s&3))||((s&3)==3))f[s]=1;
	fo(s,1,lim)for(int ss=(s-1)&s;ss;ss=(ss-1)&s)
		if((!(s&3))||(((s&3)==3)&&((!(ss&3))||((ss&3)==3)))){
			LL tmp=1;
			fo(i,1,n)if(ss&(1<<(i-1)))tmp=tmp*pw2[g[i][s-ss]]%mo;
				else if(s&(1<<(i-1)))tmp=tmp*(pw2[g[i][ss]]-1)%mo;
			f[s]=(f[s]+f[s-ss]*tmp)%mo;
		}
	printf("%d",((pw2[m]-f[lim])%mo+mo)%mo);
	return 0;
}