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>
#define ll long long
using namespace std;
const int mod=1e9+7;
inline int rd(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=0;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
	return f?x:-x;
}
int n,m,T,f[1<<15],g[16][1<<15],cnt[16][1<<15],mi[1<<15];
bool vis[16][1<<15],ok[1<<15];
int main(){
	n=rd(),m=rd();T=1<<n;mi[0]=1;
	for(int i=1;i<=n;i++) mi[i]=mi[i-1]*2;
	int x,y;
	for(int i=0;i<m;i++){
		x=rd(),y=rd();
		vis[x][1<<(y-1)]=1;
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<T;j++){
			int k=j&-j;
			cnt[i][j]=cnt[i][j^k]+vis[i][k];
			g[i][j]=mi[cnt[i][j]]-1;
		}
	}
	for(int i=0;i<T;i++) ok[i]=(((1&i)>0)==((2&i)>0));
	f[0]=1;
	for(int i=1;i<T;i++){
		if(ok[i]){
			f[i]=1;
			for(int j=(i-1)&i;j;j=(j-1)&i){
				if(ok[j]){
					int k=i^j,s=f[k];
					for(int x=0;x<n;x++){
						if(((1<<x)&k)>0){
							s=(ll)s*g[x+1][j]%mod;
						}
						else if(((1<<x)&j)>0){
							s=(ll)s*mi[cnt[x+1][k]]%mod;
						}
					}
					f[i]=(f[i]+s)%mod;
				}
			}
		}
	}
	ll ans=1;
	for(int i=1;i<=m;i++) ans=ans*2%mod;
	printf("%lld\n",(ans-f[T-1]+mod)%mod); 
	return 0;
} 
/*
5 10
2 4
3 4
2 5
2 3
1 2
3 5
1 3
1 5
4 5
1 4
*/