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
#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,b[20][20],m,A[20],B[20][20],d[20],two[110000],ans;
const int mo=1e9+7;
void get(int k,int now,int way){
//	cout<<"fa "<<k<<" "<<now<<" "<<way<<endl;
	if (k==2){
	//	for (int i=1;i<=n;i++) cout<<A[i]<<" "; cout<<endl;
	//	for (int i=0;i<=n;i++) cout<<B[1][i]<<" "; cout<<endl;
	//	for (int i=0;i<=n;i++) cout<<B[2][i]<<" "; cout<<endl;
		int rem=d[1]+d[2],tot=1; if (b[1][2]) rem--;
		for (int w=0;w<=now;w++){
			rem-=B[1][w]+B[2][w];
			ans=(ans+((1ll*way*tot%mo)<<rem))%mo;
		//	cout<<"update "<<way<<" "<<tot<<" "<<rem1<<" "<<rem2<<endl;
			if (B[1][w]==0||B[2][w]==0) break;
			tot=1ll*tot*((1<<B[1][w]+B[2][w])-(1<<B[1][w])-(1<<B[2][w])+1)%mo;
		}
		return;
	}
	int rem=d[k]; int tot=1;
	for (int w=0;w<=now;w++){
		for (int i=1;i<k;i++) if (b[i][k]) B[i][w]++; 
		rem-=B[k][w];
		get(k-1,max(w+1,now),((1ll*way*tot%mo)<<rem)%mo);
		for (int i=1;i<k;i++) if (b[i][k]) B[i][w]--;
	//	cout<<k<<" "<<w<<" "<<B[k][w]<<endl;
		if (B[k][w]==0) break;
		tot=1ll*tot*(two[B[k][w]]-1)%mo;
	}
}
int main(){
//	freopen("data.in","r",stdin);
	scanf("%d%d",&n,&m); int prem=m; two[0]=1;
	for (int i=1;i<=m;i++) two[i]=two[i-1]*2%mo;
	if (n==2){
		printf("%d\n",m); return 0;
	}
	for (;m;m--){
		int k1,k2; scanf("%d%d",&k1,&k2); b[k1][k2]=1; d[k1]++;
	}
	get(n,0,1); //cout<<ans<<" "<<two[prem]<<endl;
	ans=(two[prem]-ans+mo)%mo;
	cout<<ans<<endl;
	return 0;
}