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
/*
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20日 00時01分10秒
*/

#include<cstdio>

typedef long long ll;const int mod=1e9+7;
int dp[32768],to[15];

#define c __builtin_popcount

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


	*dp=1,m=(1<<n)-1;
	for(s=m;s;s--)
		if(dp[m^s])
			for(k=s;k;k=(k-1)&s)
				if((k&3)!=3)
				{
					v=dp[m^s];
					for(u=0;u<n;u++)
						if((s^k)&1<<u)v=(ll)v*((1<<c(k&to[u]))-1)%mod;
						else if(!(s&1<<u))v=(ll)v*(1<<c(k&to[u]))%mod;
					(dp[m^s^k]+=v)>=mod&&(dp[m^s^k]-=mod);
				}


	printf("%d\n",dp[m]);
}