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<bits/stdc++.h>
using namespace std;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	const int N=15,mod=1e9+7;
	inline int add(int x,int y){return x+y>=mod?x-mod+y:x+y;}
	inline int sub(int x,int y){return x>=y?x-y:x+mod-y;}
	inline int qpow(int x,int y){
		int ret=1;
		while(y){
			if(y&1) ret=1LL*x*ret%mod;
			x=1LL*x*x%mod,y>>=1;
		}
		return ret;
	}

	int f[1<<N],n,m,g[N+1][1<<N],e[N+1][N+1],pw[N*N];
	int main(){
		n=read<int>(),m=read<int>();
		for(int i=1;i<=m;i++){
			int u=read<int>(),v=read<int>();
			e[u][v]++;
		}
		pw[0]=1;
		for(int i=1;i<=m;i++) pw[i]=2LL*pw[i-1]%mod;

		for(int i=0;i<n;i++){
			for(int s=0;s<(1<<n);s++)
				for(int j=0;j<n;j++)
					if((s>>j)&1 && e[i+1][j+1]) g[i][s]++;
		}

		for(int S=0;S<1<<n;S++) if((S&3)==3){
			f[S]=1;
			for(int T=S;T;--T&=S) if((T&1)==(T>>1&1)){
				if(T&1){
					int ret=1;
					for(int i=0;i<n;i++) if(S>>i&1){
						if(T>>i&1) ret=1LL*ret*sub(pw[g[i][S^T]],1)%mod;
						else ret=1LL*ret*pw[g[i][T]]%mod;
					}
					f[S]=(1LL*f[T]*ret+f[S])%mod;
				}else{
					int ret=1;
					for(int i=0;i<n;i++) if(S>>i&1){
						if(T>>i&1) ret=1LL*ret*sub(pw[g[i][S^T]],1)%mod*pw[g[i][T]]%mod;
						else ret=1LL*ret*pw[g[i][T]]%mod;
					}
					f[S]=add(ret,f[S]);
				}
			}
		}
		printf("%d\n",sub(pw[m],f[(1<<n)-1]));
		return 0;
	}
}
int main(){
#ifdef my
	freopen(".in","r",stdin);
	freopen(".out","w",stdout);
#endif
	return run::main();
}