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<bits/stdc++.h>
#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=20,K=N*N,M=1e5,mo=1e9+7;

int d[N],g[N],bit[M];
int r[N];
int n,m,j,k,l,i;
ll f[M],way[K];

inline int read()
{
	int o=0; char ch=' ';
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar())o=o*10+ch-48;
	return o;
}

int main()
{
	cin>>n>>m;
	d[1]=1;	fo(i,2,n+1)d[i]=d[i-1]<<1;
	way[0]=1;
	fo(i,1,m+10)way[i]=(way[i-1]<<1)%mo;
	int T=d[n+1]-1;
	fo(i,1,T)bit[i]=bit[i>>1]+(i&1);
	fo(i,1,m){
		int x=read(),y=read();
		g[x]|=d[y];
	}
	f[0]=1;
	fo(i,3,T)if((i&1)==((i&2)>>1)){
		int u=0;
		fo(l,1,n)if(i&d[l])r[++u]=l;
		fo(l,0,d[u+1]-2){
			int o=0;
			fo(j,1,u)(l&d[j])?o|=d[r[j]]:0;
			if((o&1)!=((o&2)>>1))continue;
			int p=i^o,ok=0;
			ll choose=1;
			fo(j,1,n)if((o&d[j])>0&&(g[j]&p)==0){
				ok=1; break;
			}else (o&d[j])?choose=choose*(way[bit[g[j]&p]]-1)%mo:0;
			if(ok)continue;
			fo(j,1,n)(p&d[j])?choose=choose*way[bit[g[j]&o]]%mo:0;
			f[i]=(f[i]+choose*f[o])%mo;
		}
	}
	ll ans=(way[m]-f[T]+mo)%mo;
	cout<<ans;
}