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<bits/stdc++.h>

using namespace std;

#define N 16

const int mod=1000000007;
int n,m,ans,f[1<<N],ppc[1<<N],g[N];

void upd(int &x,int y){x=(x+y)%mod;}

int main(){
	scanf("%d%d",&n,&m);
	ans=1;
	for (int i=1;i<=m;++i){
		ans=ans*2%mod;
		int x,y; scanf("%d%d",&x,&y);
		--x; --y; g[x]|=1<<y;
	}
	for (int i=0;i<1<<n;++i)
		for (int j=0;j<n;++j) ppc[i]+=i>>j&1;
	f[0]=1;
	for (int i=1;i<1<<n;++i){
		for (int j=i;j;j=i&(j-1)){
			if ((j&1)^(j>>1&1)) continue;
			long long tmp=f[i^j];
			int k=i^j;
			for (int l=0;l<n;++l){
				if (j>>l&1){
					tmp=tmp*(1<<ppc[g[l]&k])%mod;
				}
				if (k>>l&1){
					tmp=tmp*((1<<ppc[g[l]&j])-1)%mod;
				}
			}
			upd(f[i],tmp);
		}
	}
	printf("%d\n",(ans-f[(1<<n)-1]+mod)%mod);
	
	return 0;
}