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<bits/stdc++.h>
using namespace std;
const int N=20,K=(1<<15)+10,P=1e9+7;
int g[N],num[K];
int dp[K];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		x--;y--;
		g[x]|=(1<<y);
	}
	int k=(1<<n)-1;
	for(int i=0;i<=k;i++)
		num[i]=1<<__builtin_popcount(i);
	dp[0]=1;
	for(int S=1;S<=k;S++){
		if((S&1)^(S>>1&1))continue;
		for(int U=S;U;U=(U-1)&S){//U必然不为空集
			if((U&1)^(U>>1&1))continue;
			int T=S^U,cnt=1;
			for(int p=0;p<n;p++){
				if(!(S>>p&1))continue;
				if(U>>p&1)cnt=1ll*cnt*num[g[p]&T]%P;
				else cnt=1ll*cnt*(num[g[p]&U]-1)%P;
			}
			dp[S]=(dp[S]+1ll*cnt*dp[T])%P;
		}
	}
	int all=1;
	for(int i=1;i<=m;i++)all=all*2%P;
	printf("%d\n",(all-dp[k]+P)%P);
	return 0;
}