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<iostream>
#include<cstdio>
using namespace std;
#define MOD 1000000007
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,m,G[15],bul[1<<15],bin[25],f[1<<15];
int main()
{
	n=read();m=read();
	for(int i=1,u,v;i<=m;++i)u=read()-1,v=read()-1,G[u]|=1<<v;
	int N=(1<<n)-1;f[0]=1;
	for(int i=1;i<=N;++i)bul[i]=bul[i>>1]+(i&1);
	bin[0]=1;for(int i=1;i<=n;++i)bin[i]=(bin[i-1]<<1)%MOD;
	for(int i=2;i<=N;++i)
		if((i&1)==((i>>1)&1))
			for(int U=i;U;U=(U-1)&i)
				if((U&1)==((U>>1)&1))
				{
					int T=i^U,w=1;
					for(int j=0;j<n;++j)
						if(i&(1<<j))
						{
							if(U&(1<<j))w=1ll*w*bin[bul[G[j]&T]]%MOD;
							else w=1ll*w*(bin[bul[G[j]&U]]-1)%MOD;
						}
					f[i]=(f[i]+1ll*f[T]*w)%MOD;
				}
	int ans=1;for(int i=1;i<=m;++i)ans=(ans<<1)%MOD;
	ans=(ans+MOD-f[N])%MOD;
	printf("%d\n",ans);
	return 0;
}