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;
typedef long long LL;
const int maxn=16,maxl=1<<16,mo=1e9+7;
int n,m,G[maxn],sum[maxl],mp[maxl],dp[maxl],xp[maxn*maxn];
int main()
{
	xp[0]=1;
	for (int i=1;i<maxn*maxn;i++) xp[i]=xp[i-1]*2%mo;
	cin>>n>>m;
	int ans=xp[m];
	int maxS=(1<<n)-1;
	for (int S=0;S<=maxS;S++)
	    for (int i=0;i<n;i++)
	        if (S&(1<<i))
	            mp[S]++;
	while (m--)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		u--;v--;
		G[u]|=(1<<v);
		for (int S=0;S<=maxS;S++)
		    if ((S&(1<<u))&&(S&(1<<v)))
		        sum[S]++;
	}
	for (int S=0;S<=maxS;S++)
	    if ((S&3)==3)
	    {
	    	dp[S]=1;
	        for (int U=(S-1)&S;U;U=(U-1)&S)
	        {
	        	if ((U&3)==3)
	        	{
	        		LL temp=1;
	        		for (int i=0;i<n;i++)
	        		    if (U&(1<<i))
	        		        temp=temp*xp[mp[G[i]&(S^U)]]%mo;
	        		temp=temp*xp[sum[S^U]]%mo;
	        		for (int i=0;i<n;i++)
	        		    if ((S^U)&(1<<i))
	        		        temp=temp*(xp[mp[G[i]&U]]-1)%mo;
	        		dp[S]=(dp[S]+temp)%mo;
				}
				if ((U&3)==0)
				{
	        		LL temp=1;
	        		for (int i=0;i<n;i++)
	        		    if (U&(1<<i))
	        		        temp=temp*xp[mp[G[i]&(S^U)]]%mo;
	        		temp=temp*dp[S^U]%mo;
	        		for (int i=0;i<n;i++)
	        		    if ((S^U)&(1<<i))
	        		        temp=temp*(xp[mp[G[i]&U]]-1)%mo;
	        		dp[S]=(dp[S]+temp)%mo;
				}
			}
		}
	cout<<(ans-dp[maxS]+mo)%mo;
	return 0;
}