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<cstdio>
const int MAXN=18,MAXS=(1<<15)+10,MOD=1000000007;

int N,M;
int G[MAXN];
int dp[MAXS],pw2[MAXN*MAXN];

int main()
{
	scanf("%d%d",&N,&M);
	for(int i=1;i<=M;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		v--;
		G[u]|=(1<<v);
	}
	
	dp[0]=1;
	pw2[0]=1;
	for(int i=1;i<=M;i++)
		pw2[i]=(pw2[i-1]<<1)%MOD;
	for(int S=1;S<(1<<N);S++)
		for(int U=S;U;U=(U-1)&S)
			if((U&1)==((U>>1)&1))
			{
				int T=S^U;
				int tmp=1;
				for(int i=1;i<=N;i++)
					if(S&(1<<(i-1)))
					{
						if(U&(1<<(i-1)))
							tmp=1LL*tmp*pw2[__builtin_popcount(G[i]&T)]%MOD;
						else
							tmp=1LL*tmp*(pw2[__builtin_popcount(G[i]&U)]-1)%MOD;
					}
				dp[S]=(dp[S]+1LL*dp[T]*tmp%MOD)%MOD;
			}
	int ans=(pw2[M]-dp[(1<<N)-1]+MOD)%MOD;
	printf("%d\n",ans);
	
	return 0;
}
