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>
using namespace std;
#define N 15
#define mod 1000000007

int n,m,a[N];
long long b[1<<N],f[1<<N];

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read();m=read();
	for (int i=1;i<=m;i++){
		int x=read()-1,y=read()-1;
		a[x]|=1<<y;
	}
	for (int i=b[0]=1;i<(1<<n);i++)
		b[i]=b[i/2]<<(i&1);
	for (int i=f[0]=1;i<(1<<n);i++)
		if ((i&1)==((i>>1)&1))
		for (int j=i;j;j=(j-1)&i)
			if ((j&1)==((j>>1)&1)){
				long long cnt=1;
				for (int k=0;k<n;k++){
					if (j&(1<<k)) cnt=cnt*b[a[k]&(i^j)]%mod;
					if ((i^j)&(1<<k)) cnt=cnt*(b[a[k]&j]-1)%mod;
				}
				f[i]=(f[i]+f[i^j]*cnt)%mod;
			}
	long long tmp=1;
	for (int i=1;i<=m;i++)
		tmp=tmp*2%mod;
	printf("%lld\n",(tmp-f[(1<<n)-1]+mod)%mod);
	return 0;
}