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 <cstdio>
#include <cstring>
#include <map>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
int add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
int sum[40005],n,m,S[20],st[40005];
ll f[40005];
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=40000;i++) sum[i]=sum[i-(i&-i)]+1;
	for(int i=1,x,y;i<=m;i++){
		scanf("%d%d",&x,&y);
		S[x]|=(1<<y-1);
	}
	for(int i=0;i<(1<<n);i++){
		st[i]=(1<<n)-1;
		for(int p=1;p<=n;p++) if(i&(1<<p-1)) st[i]&=S[p];
	}
	f[0]=1;
	ll ans=1;
	for(int i=1;i<=m;i++) ans=ans*2%mod;
	for(int i=1;i<(1<<n);i++){
		if((i&1)!=((i>>1)&1)) continue;
		for(int p=i;p;p=(p-1)&i){
			if((p&1)!=((p>>1)&1)) continue;
			int k=i^p;
			ll w=f[k];
			for(int j=1;j<=n;j++)
				if(k&(1<<j-1)) w=w*((1<<sum[S[j]&p])-1)%mod;
				else if(p&(1<<j-1)) w=(w<<sum[S[j]&k])%mod;
			f[i]=(f[i]+w)%mod;
		}
	}
	ans=(ans-f[(1<<n)-1])%mod;
	ans=(ans+mod)%mod;
	printf("%lld",ans);
	return 0;
}