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 rep(i,x,y) for (int i=(x);i<=(y);i++)
#define ll long long
#define inf 1000000001
using namespace std;
#define N 20
#define M 33000
#define mod 1000000007
int n,m,f[M],bin[N],a[N][M],b[N][M];
bool ok(int x){return bool(x&1)==bool(x&2);}
void upd(int &x,int y){x+=y;x-=x>=mod?mod:0;}
int main(){
	scanf("%d%d",&n,&m);
	rep (i,1,m){
		int x,y;scanf("%d%d",&x,&y);
		a[x][1<<y-1]++;
	}
	bin[0]=1;rep (i,1,n) bin[i]=2*bin[i-1];
	rep (i,1,n) rep (j,1,(1<<n)-1){
		int k=j&-j;
		b[i][j]=b[i][j^k]+a[i][k];
	}
	f[0]=1;
	rep (i,1,(1<<n)-1) if (ok(i)){
		f[i]=1;
		for (int j=i&(i-1);j;--j&=i) if (ok(j)){//j:sg必须为0
			int k=i^j,now=f[k];//k:sg大于0
			rep (l,1,n){
				if (j>>l-1&1) now=(ll)now*bin[b[l][k]]%mod;
				if (k>>l-1&1) now=(ll)now*(bin[b[l][j]]-1)%mod;
			}
			upd(f[i],now);
		}
	}
	int ans=1;
	rep (i,1,m) ans=2ll*ans%mod;
	upd(ans,mod-f[(1<<n)-1]);
	cout<<ans<<'\n';
	return 0;
}