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>
#include<algorithm>
#define MAXN 16
#define MAXM 1<<15
#define MOD 1000000007
using namespace std;
int u,v,n,bit[MAXM],cnt[MAXN][MAXM],g[MAXN][MAXM],m,bz[MAXN][MAXM],f[MAXM],ok[MAXM];
int main()
{
	scanf("%d%d",&n,&m);
	int S=1<<n;
	bit[0]=1;
	for(int i=1;i<=n;i++)
		bit[i]=bit[i-1]*2;
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&u,&v);
		bz[u][1<<v-1]=1;
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<S;j++)
		{
			int k=j&(-j);
			cnt[i][j]=cnt[i][j^k]+bz[i][k];
			g[i][j]=bit[cnt[i][j]]-1;
		}
	}
	for(int i=0;i<S;i++)
		ok[i]=(((1&i)>0)==((2&i)>0));
	f[0]=1;
	for(int i=1;i<S;i++)
		if(ok[i])
		{
			f[i]=1;
			for(int j=(i-1)&i;j;j=(j-1)&i)
				if(ok[j])
				{
					int k=i^j,s=f[k];
					for(int x=0;x<n;x++)
						if(((1<<x)&k)>0) s=1ll*s*g[x+1][j]%MOD;
						else if(((1<<x)&j)>0) s=1ll*s*bit[cnt[x+1][k]]%MOD;
						f[i]=(f[i]+s)%MOD;
				}
		}
	int ans=1;
	for(int i=0;i<m;i++)
		ans=ans*2%MOD;
	ans=(ans-f[S-1]+MOD)%MOD;
	printf("%d",ans);
}