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 <iostream>
#include <cstdio>
#include <vector>

template <typename T> void in(T &_){
	char c=getchar(); _=0; int fl=1;
	while(c<'0'||c>'9') fl=c=='-'?-1:fl, c=getchar();
	while(c>='0'&&c<='9') _=_*10+c-'0', c=getchar(); _*=fl;
}
template <typename T, typename... A> void in(T &_, A &...a){in(_); in(a...);}

#define MOD 1000000007

int N, M, bit[1<<15], f[16][1<<15], h[1<<15];
bool map[16][16];

int ksm(int x, int y){
	int z=1;
	for(; y; y>>=1, x=1LL*x*x%MOD) if(y&1) z=1LL*z*x%MOD;
	return z;
}
int lowbit(int x){return (x&(-x));}

int main(){
	for(int i=0; i<15; i++) bit[1<<i]=i+1;
	in(N, M);
	for(int x, y, i=1; i<=M; i++){
		in(x); in(y);
		map[x][y]=true;
	}
	for(int i=1; i<=N; i++)
		for(int j=0; j<(1<<N); j++)
			f[i][j]=f[i][j-lowbit(j)]+map[i][bit[lowbit(j)]];
	for(int i=0; i<(1<<N); i++){
		h[i]=1;
		for(int j=i&(i-1); j; j=i&(j-1))
			if(((j&1)^((j&2)>>1))==0 && (((i^j)&1)^(((i^j)&2)>>1))==0){
				int tmp=h[i^j];
				for(int k=j; k; k-=lowbit(k)) 
					tmp=1LL*tmp*ksm(2, f[bit[lowbit(k)]][i^j])%MOD;
				for(int k=i^j; k; k-=lowbit(k))
					tmp=1LL*tmp*(ksm(2, f[bit[lowbit(k)]][j])-1)%MOD;
				h[i]=(h[i]+tmp)%MOD;
			}
	}
	printf("%d\n", (ksm(2, M)-h[(1<<N)-1]+MOD)%MOD);
	return 0;
}