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>
#define lb(xx) (xx&(-xx))
#define MOD 1000000007
using namespace std;
inline void Add(int &a,int b)
{a+=b;a>=MOD?a-=MOD:1;}
inline int add(int a,int b)
{a+=b;return a>=MOD?a-MOD:a;}
inline int sub(int a,int b)
{a-=b;return a<0?a+MOD:a;}
inline int mul(int a,int b)
{return 1ll*a*b%MOD;}
int p2[1234];
int f[1<<16],num[1<<16][16];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	p2[0]=1;
	for(int i=1;i<=m;i++)
		p2[i]=add(p2[i-1],p2[i-1]);
	for(int i=1,x,y;i<=m;i++)
		scanf("%d%d",&x,&y),--x,--y,++num[1<<y][x];
	for(int i=0;i<(1<<n);i++)
		for(int j=0;j<n;j++)
			num[i][j]=num[lb(i)][j]+num[i^lb(i)][j];
	f[0]=1;
	for(int i=1;i<(1<<n);i++){
		if((i&1)!=(i>>1&1))continue;
		for(int j=i;j;j=(j-1)&i){
			if((j&1)!=(j>>1&1))continue;
			int pi=f[i^j],cc=0;
			for(int k=0;k<n;k++)
				if(!(i>>k&1))pi=mul(pi,p2[num[j][k]]-1);
				else if((i^j)>>k&1)cc+=num[j][k];
			Add(f[i],mul(pi,p2[cc]));
		}
	}
	printf("%d\n",sub(p2[m],f[(1<<n)-1]));
	return 0;
}