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<cstdio>
#define popcnt __builtin_popcount
typedef long long LoveLive;
const int N=15,M=32768,md=1e9+7;
int f[M],n,m,e[N];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;++i){
		int u,v;
		scanf("%d%d",&u,&v);
		e[u-1]|=1<<(v-1);
	}
	const int C=1<<n;
	f[0]=1;
	for(int S=3;S<C;++S)if((S&1)==(S>>1&1))
	for(int U=S;U;U=(U-1)&S)if((U&1)==(U>>1&1)){//sg=0
		int T=S^U;
		int T_U=1,U_T=1;
		for(int i=0;i<n;++i){
			if(U>>i&1)U_T=U_T*(1LL<<popcnt(e[i]&T))%md;else
			if(T>>i&1)T_U=T_U*((1LL<<popcnt(e[i]&U))-1)%md;
		}
		f[S]=(f[S]+(LoveLive)U_T*T_U%md*f[T])%md;
	}
	int ans=1;
	for(int i=1;i<=m;++i)ans=ans*2%md;
	ans=(ans-f[C-1]+md)%md;
	printf("%d",ans);
	return 0;
}