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;

const int mod=1000000007;
typedef long long ll;
const int N=22,N2=40001;
int f[N2],g[N],n,m;
#define pc __builtin_popcount

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=1;i<=m;++i){
		int x,y;cin>>x>>y;--x;--y;
		g[x]|=1<<y;
	}
	f[0]=1;
	for(int i=1;i<(1<<n);++i)
		if((i&1)==(i>>1&1)){
			f[i]=1;
			for(int j=(i-1)&i;j;j=(j-1)&i){
				if((j&1)!=(j>>1&1))continue;
				int res=f[i-j];
				for(int k=0;k<n;++k)
					if(i>>k&1){
						if(j>>k&1)res=((ll)res<<pc(g[k]&(i-j)))%mod;
						else res=(((ll)res<<pc(g[k]&j))-res)%mod;
					}
				f[i]=(f[i]+res)%mod;
			}
		}
	int ans=1;
	for(int i=1;i<=m;++i)ans=ans*2%mod;
	ans=(ans-f[(1<<n)-1]+mod)%mod;
	cout<<ans<<'\n';
	return 0;
}