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;
typedef long long ll;
const int mod=1e9+7;
const int N=40000;
int n,m,u,v,zt[20],f[N],pw[N];
void up(int &x,int y){x=x-mod+y; x=(x<0)?x+mod:x;}
bool check(int s)
{
	return (s&1)&&((s>>1)&1)||!(s&1)&&!((s>>1)&1);
}
int count(int s)
{
	int cnt=0;
	while (s) cnt++,s-=(s&(-s));
	return cnt;
}
int main()
{
	scanf("%d%d",&n,&m);
	pw[0]=1;
	for (int i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		zt[u]|=(1<<(v-1));//保存出边状态 
		pw[i]=(ll)pw[i-1]*2%mod;
	}
	int Max=1<<n;
	f[0]=1;
	for (int i=0;i<Max;i++)
	if (f[i])
	{
		int _i=(Max-1)^i;
		for (int j=_i;j;j=((j-1)&_i))
		if (check(j)) 
		{
			int del=1;
			for (int k=1;k<=n;k++)
			{
				if ((i>>(k-1))&1) 
				  del=(ll)del*(pw[count(zt[k]&j)]-1+mod)%mod;
			    if ((j>>(k-1))&1)
			      del=(ll)del*pw[count(zt[k]&i)]%mod;
			}
			up(f[i|j],(ll)del*f[i]%mod);
		}
	}
	printf("%d\n",((ll)pw[m]-f[Max-1]+mod)%mod);
	return 0;
} 