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 <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int N=16,M=1<<15,mo=1e9+7;

typedef long long LL;

int n,m,T,f[M],g[N][M],cnt[N][M],p[M];

bool bz[N][M],ok[M];

int main()
{
	scanf("%d%d",&n,&m); T=1<<n;
	p[0]=1;
	for (int i=1;i<=n;i++) p[i]=p[i-1]*2;
	for (int i=0,x,y;i<m;i++)
	{
		scanf("%d%d",&x,&y);
		bz[x][1<<y-1]=1;
	}
	for (int i=1;i<=n;i++)
	{
		for (int j=1;j<T;j++)
		{
			int k=j&-j;
			cnt[i][j]=cnt[i][j^k]+bz[i][k]; g[i][j]=p[cnt[i][j]]-1;
		}
	}
	for (int i=0;i<T;i++) ok[i]=(((1&i)>0)==((2&i)>0));
	f[0]=1;
	for (int i=1;i<T;i++) if (ok[i])
	{
		f[i]=1;
		for (int j=(i-1)&i;j;j=(j-1)&i) if (ok[j])
		{
			int k=i^j,s=f[k];
			for (int x=0;x<n;x++) if (((1<<x)&k)>0) s=(LL)s*g[x+1][j]%mo;
			else if (((1<<x)&j)>0) s=(LL)s*p[cnt[x+1][k]]%mo;
			f[i]=(f[i]+s)%mo;
		}
	}
	int ans=1;
	for (int i=0;i<m;i++) ans=ans*2%mo;
	ans=(ans-f[T-1]+mo)%mo;
	printf("%d\n",ans);
	return 0;
}