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<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#define LL long long
using namespace std;
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int N=(1<<15)+3;
const int mod=1000000007;
int n,m,f[N],num[N],e[16];
inline void up(int &x,int y) {x=x+y>=mod?x+y-mod:x+y;}
int main( )
{
  int i,j,x,y;
  n=read( );m=read( );
  for(i=1;i<=m;i++)
    x=read( ),y=read( ),e[x-1]|=(1<<(y-1));
  int S=(1<<n)-1,B,k,u;LL r;
  num[0]=0;f[0]=1;
  for(i=1;i<=S;i++) 
    num[i]=num[i>>1]+(i&1);
  for(i=0;i<S;i++)
    if(x=f[i])
      for(B=S^i,k=B;k;k=(k-1)&B)
	if((k&3)!=3)
	  {
	    r=x;u=B^k;
	    for(j=0;j<n;j++)
	      if((u>>j)&1) r=r*((1<<num[e[j]&k])-1)%mod;
	      else if((i>>j)&1) r=r*(1<<num[e[j]&k])%mod;
	    up(f[i|k],r);
	  }
  printf("%d",f[S]);
  return 0;
}