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<cstdio>
#include<cstring>
#include<algorithm>
#define N 32770
#define ll long long
#define mod 1000000007
using namespace std;
int n, m, bin[400], map[20][20], x, y, w1[20][N], w2[20][N], f[N];
int main(){
	scanf("%d%d", &n, &m);
	memset(map, 0, sizeof(map));
	for(int i=1; i<=m; i++){scanf("%d%d", &x, &y); map[x][y]=1;}
	bin[0]=1; for(int i=1; i<=n*n; i++)bin[i]=bin[i-1]*2%mod;
	for(int i=1; i<=n; i++)
		for(int j=0; j<=bin[n]-1; j++){
			w1[i][j]=w2[i][j]=0;
			for(int k=1; k<=n; k++)if(j&bin[k-1]){
				w1[i][j]+=map[i][k]; w2[i][j]+=map[k][i];
			}
		}
	f[0]=1;
	for(int i=1; i<=bin[n]-1; i++){
		f[i]=0;
		for(int j=(i-1)&i; j>=0; j=(j-1)&i){
			if(j&bin[0]&&!(j&bin[1]))continue;
			if(!(j&bin[0])&&j&bin[1])continue;
			x=1;
			for(int k=1; k<=n; k++)if(j&bin[k-1])x=(ll)x*(bin[w1[k][i-j]]-1)%mod*bin[w2[k][i-j]]%mod;
			f[i]=(f[i]+(ll)f[j]*x%mod)%mod;
			if(!j)break;
		}
	}
	printf("%d", (bin[m]-f[bin[n]-1]+mod)%mod);
	return 0;
}