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:
/*
srt: 2017年06月19日 22時00分00秒
   《将赴吴兴登乐游原》杜牧
清时有味是无能，闲爱孤云静爱僧。
欲把一麾江海去，乐游原上望昭陵。
end: 2017年06月19日 23時27分31秒
*/

#include<cstdio>

typedef long long ll;const int mod=1e9+7;
int dp[15][32768],f[14348907],g[14348907],b[32768],to[15],n;

#define c __builtin_popcount

void prework()
{
	int s,k,t,r,u;b[1]=1;
	for(k=1;k<n;k++)b[1<<k]=b[1<<(k-1)]*3;
	for(s=1;s<1<<n;s++)
	{
		if(c(s)>1)t=s&-s,b[s]=b[s^t]+b[t];
		s[b][f]=1;
		for(k=s;k;k=(k-1)&s)
		{
			t=r=1;
			for(u=0;u<n;u++)
				if((s^k)&1<<u)
					r=(ll)r*(1<<c(k&to[u]))%mod,
					t=(ll)t*((1<<c(k&to[u]))-1)%mod;
			f[b[s]+b[k]]=r;
			g[b[s]+b[k]]=t;
		}
	}
}

int main()
{
	int i,s,k,u,v,m;scanf("%d%d",&n,&m);
	for(i=0;i<m;i++)scanf("%d%d",&u,&v),to[--u]|=1<<--v;

	prework();m=(1<<n)-1;

	for(i=0;i<n;i++)
		for(s=0;s<1<<n;s++)
			if(i||s==m)
				for(k=s;k;k=(k-1)&s)
					if((k&3)!=3)
						dp[i][m^s^k]+=(i?(ll)dp[i-1][m^s]:1ll)*
							f[b[s]+b[s^k]]%mod*
							g[b[s]+b[k]]%mod,dp[i][m^s^k]>=mod&&(dp[i][m^s^k]-=mod);

	s=0;
	for(i=1;i<n;i++)(s+=dp[i][m])>=mod&&(s-=mod);
	printf("%d\n",s);
}