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<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 110
#define MAXM 32800
#define MO 1000000007
int f[MAXM],pow2[MAXN],n,m,tot,G[20];
int Count(int x)
{
	int ret=0;
	while(x)
	{
		ret+=x&1;
		x>>=1;
	}
	return ret;
}
int main()
{
	int x,y;
	scanf("%d%d",&n,&m);
	tot=1<<n;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		x--,y--;
		G[x]|=(1<<y);
	}
	f[0]=1;pow2[0]=1;
	for(int i=1;i<=m;i++)
		pow2[i]=2LL*pow2[i-1]%MO;
	for(int S=1;S<tot;S++)
		if((S&1)==((S>>1)&1))
		{
			for(int U=S;U;U=(U-1)&S)
				if((U&1)==((U>>1)&1))
				{
					int w=1;
					for(int i=0;i<n;i++)
						if((S>>i)&1)
						{
							if((U>>i)&1) w=1LL*w*pow2[Count(G[i]&(S^U))]%MO;
							else w=1LL*w*(pow2[Count(G[i]&U)]-1)%MO;
						}
					f[S]=(f[S]+1LL*w*f[S^U]%MO)%MO;
				}
		}
	int ans=((pow2[m]-f[tot-1])%MO+MO)%MO;
	printf("%d\n",ans);
}