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<bits/stdc++.h>
using namespace std;
#define LL long long

const int mo=1e9+7;
int n,m,g[15],pow2[233],bit[1<<15],dp[1<<15];

int main()
{
#ifdef h10
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	int i,u,v,S,S0;
	scanf("%d%d",&n,&m);
	pow2[0]=1;
	for (i=1;i<233;i++)
		pow2[i]=2*pow2[i-1]%mo;
	for (i=1;i<1<<n;i++)
		bit[i]=bit[i>>1]+(i&1);
	for (i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		u--; v--;
		g[u]|=1<<v;
	}

	dp[0]=1;
	for (S=1;S<1<<n;S++)
	{
		if ((S&1)^((S&2)>>1)) continue;
		for (S0=S;S0;S0=(S0-1)&S)
		{
			int Sum=dp[S-S0];
			if (!Sum) continue;
			for (i=0;i<n;i++)
				if ((S>>i)&1)
				{
					if ((S0>>i)&1)
						Sum=(LL)Sum*pow2[bit[g[i]&(S-S0)]]%mo;
					else
						Sum=(LL)Sum*(pow2[bit[g[i]&S0]]-1)%mo;
				}
			(dp[S]+=Sum)%=mo;
		}
	}
	printf("%d\n",(pow2[m]+mo-dp[(1<<n)-1])%mo);
}
