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<cstdio>
#include<algorithm>
using namespace std;
const long long md=1e9+7;
int w[20];
long long pow2[401];
long long dp[1<<15];
int main(){
//	freopen("F.in","r",stdin);
	pow2[0]=1;
	for(int i=1;i<401;i++)
		pow2[i]=pow2[i-1]*2%md;
	int n,m;
	scanf("%d%d",&n,&m);
	int s,t;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&s,&t);
		s--,t--;
		w[s]|=(1<<t);
	}
	dp[0]=1;
	long long st1,st2;
	for(int i=1;i<(1<<n);i++)
	{
		dp[i]=1;
		for(int j=i&(i-1);j;j=i&(j-1))
		{
			s=i^j;
			if((j&1)!=((j&2)>>1)) continue;
			st1=1,st2=1;
			for(int k=0;k<n;k++)
			{
				if(s&(1<<k)) st1=st1*pow2[__builtin_popcount(w[k]&j)]%md;
				if(j&(1<<k)) st2=st2*(pow2[__builtin_popcount(w[k]&s)]-1)%md;
			}
			dp[i]=(dp[i]+dp[j]*st1%md*st2)%md;
		}
	}
	printf("%lld\n",(pow2[m]-dp[(1<<n)-1]+md)%md);
	return 0;
}