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<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
int n,m,shu1,shu2,point[1000005],dp[1000005];
int ans,num[1000005];
int moc(int x)
{
	if(x>=mod) 
        return x-mod;
	if(x<0) 
        return x+mod;
	return x;
}
int main()
{
	cin>>n>>m;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&shu1,&shu2);
		shu1--,shu2--;
		point[shu1]|=1<<shu2;
	}
	int maxa=(1<<n)-1;
	dp[0]=1;
	for(int i=0;i<=maxa;i++)
	{
		int now=0;
		for(int j=0;j<n;j++)
		    if(i>>j&1) 
                now++;
		num[i]=1<<now;
	}
	for(int s=1;s<=maxa;s++)
	if((s&1)==(s>>1&1))
	{
		for(int ss=s;ss;ss=(ss-1)&s)
        {
            if((ss&1)==(ss>>1&1))
		    {
			    int t=s^ss,now=1;
			    for(int i=0;i<n;i++)
			    if(ss>>i&1)
			        now=1ll*now*num[point[i]&t]%mod;
			    else if(t>>i&1)now=1ll*now*(num[point[i]&ss]-1)%mod;
			        dp[s]=moc(dp[s]+1ll*now*dp[t]%mod);
		    }
        }	    
	}
	ans=1;
	for(int i=1;i<=m;i++)
	    ans=moc(ans<<1);
	ans=moc(ans-dp[maxa]);
	printf("%d",ans);
	return 0;
}