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<iostream>
#include<unordered_map>
#include<cstdio>
using namespace std;
#define N 16
#define U 65537
#define mod 1000000007
inline int M(int x){return (x>=mod)?(x-mod):x;}
inline int M1(int x){return (x<0)?(x+mod):x;}
inline int ksm(int p,int k){int res=1;while(k){if(k&1)res=1ll*res*p%mod;k>>=1;p=1ll*p*p%mod;}return res;}
int n,m,all,ans,f[U],w[N][N],deg[N][U],P[N];unordered_map<int,int>g[U];
int main(){
	P[0]=0;for(int i=1;i<N;i++)P[i]=M(P[i-1]<<1|1);
	scanf("%d%d",&n,&m);all=(1<<n)-1;int i,x,y,res,S,T,p,v;
	for(i=1;i<=m;i++){scanf("%d%d",&x,&y);++w[x-1][y-1];}
	for(x=0;x<n;x++)
	for(S=0;S<=all;S++)
	for(y=0;y<n;y++)
	if((S&(1<<y))&&w[x][y])deg[x][S]+=w[x][y];
	ans=ksm(2,m);f[0]=1;
	for(T=3;T<=all;T++)
	if(((T&3)==3)||((T&3)==0))
	for(S=T;S;S=T&(S-1))
	{
		p=T^S;v=1;
		if(((p&3)==3)||((p&3)==0))
		{
			for(x=0;x<n;x++)
			if(T&(1<<x))
			{
				if(p&(1<<x))v=1ll*v*P[deg[x][S]]%mod;
				else v=1ll*v*M(P[deg[x][p]]+1)%mod;
			}
		}f[T]=(f[T]+1ll*f[p]*v)%mod;
	}printf("%d",M1(ans-f[all]));return 0;
}