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<bits/stdc++.h>
#define ll long long
#define mod 1000000007
#define N 19
#define E 259
#define M 40009
using namespace std;

int n,m,tot,lg2[E],num[M],a[N],f[M][N],g[2][M][E],h[2][M][E],dp[M];
int calc(int x,int y){
	return (ll)g[0][y][x&255]*g[1][y][x>>8]%mod;
}
int calceg(int x,int y){
	return (ll)h[0][y][x&255]*h[1][y][x>>8]%mod;
}
int main(){
	scanf("%d%d",&n,&m);
	int i,j,k,l,x,y;
	for (i=1; i<=m; i++){
		scanf("%d%d",&x,&y); a[x]|=1<<y-1;
	}
	tot=1<<n;
	for (i=0; i<=8; i++) lg2[1<<i]=i;
	for (i=num[0]=1; i<tot; i++) num[i]=num[i^i&-i]<<1;
	for (i=0; i<tot; i++){
		for (j=1; j<=16; j++) f[i][j]=num[a[j]&i];
		for (j=g[0][i][0]=g[1][i][0]=h[0][i][0]=h[1][i][0]=1; j<256; j++){
			k=lg2[j&-j]+1; l=j^j&-j;
			g[0][i][j]=(ll)g[0][i][l]*(f[i][k]-1)%mod;
			g[1][i][j]=(ll)g[1][i][l]*(f[i][k+8]-1)%mod;
			h[0][i][j]=(ll)h[0][i][l]*f[i][k]%mod;
			h[1][i][j]=(ll)h[1][i][l]*f[i][k+8]%mod;
		}
	}
	dp[0]=1;
	for (i=1; i<tot; i++)
		for (j=i; j; j=(j-1)&i) if ((j&3)<3)
			dp[i]=(dp[i]+(ll)dp[i^j]*calc(i^j,j)%mod*calceg(j,i^j))%mod;
	printf("%d\n",dp[tot-1]);
	return 0;
}
