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> 
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
typedef long long ll;
using namespace std;
const int mo=1e9+7,maxn=(1<<15)+5;
int n,m,i,j,k,s,num[maxn],cnt[maxn],g[maxn],p[16][maxn];
int f[maxn],two[400],ans;
int main(){
	scanf("%d%d",&n,&m);
	two[0]=1;
	fo(i,1,m){
		int a,b;
		scanf("%d%d",&a,&b);
		g[a]|=1<<b-1;
		two[i]=two[i-1]*2%mo;
	}ans=two[m];
	m=(1<<n)-1;
	fo(i,1,m) num[i]=num[i>>1]+(i&1);
	fo(j,1,n){
		s=g[j];
		fo(i,0,m) p[j][i]=num[i&s];
	}
	fo(i,0,m){
		int cn=0;
		fo(j,1,n) if (i&(1<<j-1)) cn+=p[j][i];
		cnt[i]=cn;
	}
	fo(i,0,m)if ((i&3)==3){
		int sum=1,an,t;
		for(j=i;j;j=(j-1)&i){
			s=i^j,an=1,t=0;
			if (!((s&3)==3||(j&3)==3)) continue;
			fo(k,0,n-1){
				if (j&(1<<k)) an=(ll)((1<<p[k+1][s])-1)*an%mo;	
				if (s&(1<<k)) t+=p[k+1][j];
			}
			an=(ll)an*two[t]%mo;
			if (j&3) (sum+=(ll)an*f[j]%mo)%=mo;else
			(sum+=(ll)an*two[cnt[j]]%mo)%=mo;
		}f[i]=sum;
	}
	printf("%d\n",(ans-f[m]+mo)%mo);
}