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;
#define ll long long
const int maxn=16,mod=1e9+7;
int qpow(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)ans=(ll)ans*a%mod;
		a=(ll)a*a%mod;
		b>>=1;
	}
	return ans;
}
int add(int a,int b)
{
	a+=b;return a>=mod?a-mod:a;
}
int p2[maxn*maxn]={1},dp[1<<maxn|10];
int n,m,yc[maxn+1][1<<maxn|10];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)p2[i]=add(p2[i-1],p2[i-1]);
    for(int i=1;i<=m;i++)
    {
    	int u,v;
    	scanf("%d%d",&u,&v);
    	for(int s=0;s<(1<<n);s++)
    	{
    		if((s>>v-1)&1)
    		{
    			yc[u][s]++;
			}
		}
	}
	for(int s=0;s<(1<<n);s++)
	{
		if(((s>>0)&1)==((s>>1)&1))
		{
		    for(int u=s;u;u=(u-1)&s)
		    {
		    	if(((u>>0)&1)==((u>>1)&1))
		    	{
		    		int v=s-u;
		    		int s1=0,s2=1;
		    		for(int i=1;i<=n;i++)
		    		{
		    			if((v>>i-1)&1)
		    			{
		    				s1=add(s1,yc[i][u]);
						}
						if((u>>i-1)&1)
						{
							s2=(ll)s2*(p2[yc[i][v]]-1)%mod;
						}
					}
					dp[s]=add(dp[s],(ll)dp[u]*s2%mod*p2[s1]%mod);
				}
			}
		}
		dp[s]=add(dp[s],1);
	}
	cout<<add(p2[m],mod-dp[(1<<n)-1]);
	return 0;
}