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 <cstdio>
#include <cstring>
#include <vector>
#define MAXN 20
#define MAXS 33000
#define LL long long
using namespace std;

const LL P=1000000007;

int n,m,maxs;
int w[MAXN][MAXN];
LL f[MAXN][MAXS];
vector< pair<int,LL> > trans[MAXS];
LL g[MAXS];

void pre_gao(int x){
	for(int y=x;;y=(y-1)&x){
		LL res=1;
		int cnt=0;
		for(int i=1;i<=n;i++)
			if((1<<(i-1))&y){
				int temp=0;
				for(int j=i+1;j<=n;j++)
					if(((1<<(j-1))&x) && !((1<<(j-1))&y) && w[i][j])
						temp++;
				res=res*((1LL<<temp)-1)%P;
			}else if((1<<(i-1))&x){
				for(int j=i+1;j<=n;j++)
					if(((1<<(j-1))&y) && w[i][j])
						cnt++;
			}
		while(cnt--) res=res*2%P;
		if(res) trans[x].push_back(make_pair(y,res));
		if(!y) break;
	}
	int temp=0;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			if(w[i][j] && ((1<<(i-1))&x) && ((1<<(j-1))&x))
				temp++;
	g[x]=1;
	while(temp--) g[x]=g[x]*2%P;
}

inline void update(LL &x,LL y){ x=(x+y)%P; }

int main(){
#ifdef DEBUG
	freopen("F.in","r",stdin);
#endif
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		w[x][y]=1;
	}
	maxs=1<<n;
	for(int i=0;i<maxs;i++) pre_gao(i);
	f[0][maxs-1]=1;
	LL ans=0;
	for(int i=0;i<n;i++)
		for(int j=0;j<maxs;j++)
			if(f[i][j]){
				for(int k=0;k<trans[j].size();k++)
					if((trans[j][k].first&3)==3)
						update(f[i+1][trans[j][k].first],f[i][j]*trans[j][k].second);
					else if((trans[j][k].first&3)==0)
						update(ans,f[i][j]*trans[j][k].second%P*g[trans[j][k].first]);
			}
	LL temp=1;
	for(int i=1;i<=m;i++) temp=temp*2%P;
	ans=(temp-ans+P)%P;
	printf("%lld\n",ans);
}