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>
using namespace std;
#define N 16
#define ll long long
const ll mod=1e9+7;

int n,m,x,y;
int s[N][1<<N],ok[1<<N];
ll g[N][1<<N],f[1<<N],p[N];

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++) {
		scanf("%d%d",&x,&y); x--; y--;
		s[x][1<<y]=1;
	}
	for (int i=0;i<n;i++)
		for (int j=0;j<(1<<n);j++) {
			int k=j&(-j);
			g[i][j]=g[i][j-k]+s[i][k];
		}
	for (int i=0;i<(1<<n);i++) if ((i&1)==((i&2)>>1)) ok[i]=1;
	p[0]=1; for (int i=1;i<N;i++) p[i]=p[i-1]*2%mod;
	
	f[0]=1;
	for (int i=1;i<(1<<n);i++) if (ok[i]) {
		f[i]=1;
		for (int j=(i-1)&i;j;j=(j-1)&i) if (ok[j]) {
			ll tmp=f[i-j];
			for (int k=0;k<n;k++)
				if (j>>k&1) tmp=tmp*p[g[k][i-j]]%mod; else
				if ((i-j)>>k&1) tmp=tmp*(p[g[k][j]]-1+mod)%mod;
			f[i]=(f[i]+tmp)%mod;
		}
	}
	ll ans=1; for (int i=1;i<=m;i++) ans=ans*2%mod;
	printf("%lld\n",(ans-f[(1<<n)-1]+mod)%mod);
	return 0;
}