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<algorithm>
#include<cstring>
#define ll long long
using namespace std;
const int N=1<<15|10,P=1e9+7;
int n,m,U,e[20],pw[300],cnt[N],f[N];
bool can[N];

inline void read(int &i){
	i=0;char c=getchar();bool j=0;
	for(;!isdigit(c);c=getchar())if(c=='-')j=1;
	for(;isdigit(c);c=getchar())i=(i<<1)+(i<<3)+c-'0';i=j?-i:i;
}
inline void add(int &x,int y){(x+=y)>=P?x-=P:0;}

int main(){
	read(n);read(m);U=(1<<n)-1;pw[0]=1;
	for(int i=1,x,y;i<=m;++i){
		read(x);read(y);
		x--;y--;
		e[x]|=1<<y;
	}
	for(int i=1;i<=U;++i)cnt[i]=cnt[i>>1]+(i&1);
	for(int i=1;i<=m;++i)pw[i]=pw[i-1],add(pw[i],pw[i-1]);
	for(int s=0;s<=U;++s)can[s]=(s&1)==(s>>1&1);
	f[0]=1;
	for(int s=1;s<=U;++s)if(can[s])
		for(int t=s;t;t=(t-1)&s)if(can[t]){
			ll as=1;
			for(int i=0;i<n;++i)if(s>>i&1){
				if(t>>i&1)as=as*pw[cnt[e[i]&(s^t)]]%P;
				else as=as*(pw[cnt[e[i]&t]]-1)%P;
			}
			add(f[s],as*f[s^t]%P);
		}
	printf("%d\n",(pw[m]-f[U]+P)%P);
	return 0;
}