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<cstdio>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const int mo=1000000007;
bool bz[15+5][15+5];
int two[300],id[1<<17];
int f[15+5][1<<17],g[15+5][1<<17],dp[1<<17];
int i,j,k,l,r,s,t,n,m,ans;
int lowbit(int x){
	return x&-x;
}
int main(){
	scanf("%d%d",&n,&m);
	fo(i,1,n) id[1<<(i-1)]=i;
	two[0]=1;
	fo(i,1,m) two[i]=(ll)two[i-1]*2%mo;
	fo(i,1,m){
		scanf("%d%d",&j,&k);
		bz[j][k]=1;
	}
	fo(i,1,n)
		fo(s,1,(1<<n)-1)
			f[i][s]=f[i][s-lowbit(s)]+bz[i][id[lowbit(s)]];
	fo(i,1,n)
		fo(s,1,(1<<n)-1){
			k=id[lowbit(s)];
			if (!bz[i][k]) g[i][s]=g[i][s-lowbit(s)];
			else g[i][s]=((ll)g[i][s-lowbit(s)]*2%mo+1)%mo;
		}
	fo(s,1,(1<<n)-1){
		//if (!((s&1)&&(s&2))) continue;
		dp[s]=1;
		t=(s-1)&s;
		while (t){
			if (((s^t)&1)&&(t&2)){
				t=(t-1)&s;
				continue;
			}
			if (((s^t)&2)&&(t&1)){
				t=(t-1)&s;
				continue;
			}
			r=1;
			fo(i,1,n)
				if ((t&(1<<(i-1)))) r=(ll)r*g[i][s^t]%mo;
			l=0;
			fo(i,1,n)
				if (((s^t)&(1<<(i-1)))) l+=f[i][t];
			r=(ll)r*two[l]%mo;
			(dp[s]+=(ll)dp[t]*r%mo)%=mo;
			t=(t-1)&s;
		}
	}
	ans=(two[m]-dp[(1<<n)-1])%mo;
	(ans+=mo)%=mo;
	printf("%d\n",ans);
}