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>
#define N 100050
using namespace std;
const int mod=1e9+7;
int n,m,ed[20],dp[N],cnt[N],bin[30];
inline void modd(int &x){x-=x>=mod?mod:0;}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<=20;++i)bin[i]=1<<i;
	for(int i=0;i<bin[n];++i)cnt[i]=cnt[i>>1]+(i&1);
	for(int i=1,x,y;i<=m;++i)
		scanf("%d%d",&x,&y),ed[x-1]|=bin[y-1];
	int ans=0;
	dp[bin[n]-1]=1;
	for(int i=bin[n]-1,fst;i;--i){
		fst=bin[n]-1-i;
		for(int j=fst;j;j=j-1&fst)if(j&3^3){
			int p=1;
			for(int k=0;k<n;++k)if(i>>k&1){
				p=1ll*p*(bin[cnt[ed[k]&j]]-1)%mod;
			}
			else if(j>>k&1){
				p=1ll*p*bin[cnt[ed[k]&i]]%mod;
			}
			dp[i]=(1ll*dp[i|j]*p+dp[i])%mod;
		}
//		printf("dp[%d]=%d\n",i,dp[i]);
		if(i&3^3)ans+=dp[i],modd(ans);
	}
	cout<<ans<<endl;
	return 0;
}