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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
int f[1<<15],p[1<<15];
int main()
{
	const int mod=1e9+7;
	int i,k,s,t,o,n=gi(),m=gi(),e[15]={},w;
	const int all=(1<<n)-1;
	while (m--) k=gi()-1,e[k]|=1<<(gi()-1);
	for (i=p[0]=1;i<1<<n;i++) p[i]=(p[i^(i&-i)]<<1)%mod;
	
	f[0]=1;
	for (s=0;s<1<<n;s++)
		if (f[s])
			for (o=s^all,t=o;t;(--t)&=o) {
				if ((t&3)==3) continue;
				w=f[s];
				for (i=0;i<n;i++)
					if (s>>i&1) {
						if (!(e[i]&t)) break;
						w=1LL*w*(p[e[i]&t]-1)%mod;
					} else if (t>>i&1) w=1LL*w*p[e[i]&s]%mod;
				if (i==n) (f[s|t]+=w)%=mod;
			}
	printf("%d\n",f[all]);
	return 0;
}
