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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int N=20;
const int S=1<<N;
const int mod=1000000007;

int n,sz,x,y,m,ans;
int dp[S],g[N],cnt[S],bin[S];

void init(){
	int i,j;
	for (i=0;i<=sz;i++)
		for (j=0;j<n;j++) if ((i>>j)&1) cnt[i]++;
}

int main(){
	int i,s,ss,u,w;
	bin[0]=1; for (i=1;i<S;i++) bin[i]=2ll*bin[i-1]%mod;
	scanf("%d%d",&n,&m); sz=(1<<n)-1;
	for (i=1;i<=m;i++){
		scanf("%d%d",&x,&y); x--; y--;
		g[x]|=1<<y;
	}
	init();
	dp[0]=1;
	for (s=1;s<=sz;s++) if ((s&1)==((s>>1)&1)){
		for (ss=s;ss;ss=(ss-1)&s) if ((ss&1)==((ss>>1)&1)){
			u=ss^s; w=1;
			for (i=0;i<n;i++) if ((s>>i)&1){
				if ((u>>i)&1) w=1ll*w*(bin[cnt[g[i]&ss]]-1)%mod;
				else w=1ll*w*bin[cnt[g[i]&u]]%mod;
			}
			(dp[s]+=1ll*dp[u]*w%mod)%=mod;
		}
	}
	ans=(bin[m]-dp[sz]+mod)%mod;
	printf("%d\n",ans);
	return 0;
}