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<set>
#include<map>
#include<cmath>
#include<queue>
#include<bitset>
#include<string>
#include<cstdio>
#include<cctype>
#include<cassert>
#include<cstdlib>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>

#define For(i,x,y) for (int i=x;i<y;i++)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define lf else if

#define dprintf(...) fprintf(stderr,__VA_ARGS__)
using namespace std;

typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> Vi;

int IN(){
	int c,f,x;
	while (!isdigit(c=getchar())&&c!='-');c=='-'?(f=1,x=0):(f=0,x=c-'0');
	while (isdigit(c=getchar())) x=(x<<1)+(x<<3)+c-'0';return !f?x:-x;
}

const int N=15;
const int p=1e9+7;

int Pow(int a,int b){
	int res=1;
	for (;b;b>>=1,a=1ll*a*a%p) if (b&1) res=1ll*res*a%p;
	return res;
}

int g[N][1<<N],w[N][1<<N],f[1<<N],chk[1<<N];
int A[N][N];
int n,m,x,y;

int main(){
	n=IN();m=IN();
	For(i,0,m){
		x=IN()-1,y=IN()-1;
		A[x][y]=1;
	}
	For(i,0,n) For(t,0,1<<n)
		if (!(t>>i&1)){
			int c=0;
			For(j,0,n) if ((t>>j&1)&&A[i][j]) c++;
			g[i][t]=(1<<c)-1;
			w[i][t]=1<<c;
		}
	For(t,1,1<<n) chk[t]=((t&1)==(t>>1&1));
	For(s,1,1<<n) if (chk[s]){
		f[s]=1;
		for (int u=(s-1)&s;u;u=(u-1)&s)
			if (chk[u]&&chk[s^u]){
				int tmp=1;
				For(i,0,n)
					if (u>>i&1){
						tmp=1ll*tmp*g[i][s^u]%p;
					} lf (s>>i&1){
						tmp=1ll*tmp*w[i][u]%p;
					}
				f[s]=(f[s]+1ll*tmp*f[u])%p;
			}
	}
	printf("%d\n",(Pow(2,m)-f[(1<<n)-1]+p)%p);
}