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<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
const int MAXN=15;
const int MAXS=(1<<MAXN);
const int MOD=1000000007;
int n,m,S,tot=1;
int f[MAXS+5]={1},cnt[MAXS+5],to[MAXN+10];
int main()
{
	scanf("%d%d",&n,&m);S=(1<<n)-1;
	for(int i=0,u,v;i<m;++i)
	{
		scanf("%d%d",&u,&v);
		to[--u]^=(1<<(--v));
		tot=tot+tot;if(tot>=MOD)tot-=MOD;
	}
	for(int i=S;~i;--i)for(int t=i;t;t^=t&(-t))++cnt[i];
	for(int i=0,la,x,y,w;i<S;++i)
	{
		la=S^i;if((i&3)&&(i&3)!=3)continue;
		for(int j=la;j;j=(j-1)&la)
		{
			x=i^j;if((x&3)&&(x&3)!=3)continue;
			y=S^x;w=f[i];
			for(int o=0;o<n;++o)
				if((y>>o)&1)w=(ll)w*((1<<cnt[j&to[o]])-1)%MOD;
				else if((j>>o)&1)w=((ll)w<<cnt[y&to[o]])%MOD;
			f[x]=f[x]+w;if(f[x]>=MOD)f[x]-=MOD;
		}
	}
	tot=tot+MOD-f[S];
	printf("%d",(tot>=MOD)?(tot-MOD):tot);
	return 0;
}