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>
typedef long long lint;
const int N=15;
const int mod=1e9+7;
int n,m,nxt[N],dp[1<<N],cbit[1<<N];

inline int nxi(){
	int x=0;
	char c;
	while((c=getchar())>'9'||c<'0');
	while(x=x*10-48+c,(c=getchar())>='0'&&c<='9');
	return x;
}

template <class T> inline int fpow(int x,T t){
	int ans=1;
	for(; t; x=(lint)x*x%mod,t>>=1){
		if(t&1) ans=(lint)ans*x%mod;
	}
	return ans;
}

int main(){
	n=nxi(),m=nxi();
	for(int i=1; i<=m; ++i){
		const int a=nxi()-1,b=nxi()-1;
		nxt[a]|=1<<b;
	}
	for(int i=1; i<1<<n; ++i){
		cbit[i]=cbit[i>>1]+(i&1);
	}
	dp[3]=1;
	for(int i=4; i<1<<n; ++i){
		if((i&1)!=(i>>1&1)) continue;
		dp[i]=1;
		for(int j=(i-1)&i; j; j=(j-1)&i){
			if((j&1)!=(j>>1&1)) continue;
			const int p=i^j;
			int ws=dp[j],wt=1;
			for(int k=0; k<n&&ws; ++k){
				if(j>>k&1){
					ws=ws*((1ll<<cbit[nxt[k]&p])-1)%mod;
				}
				if(p>>k&1){
					wt=wt*(1ll<<cbit[nxt[k]&j])%mod;
				}
			}
			dp[i]=(dp[i]+(lint)ws*wt)%mod;
		}
	}
	printf("%d\n",(fpow(2,m)-dp[(1<<n)-1]+mod)%mod);
	return 0;
}