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:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define mxn (1<<16)
#define ll long long
#define lowbit(x) (x&-x)
#define mdn 1000000007
using namespace std;

int f[mxn],in[16],n,m;
int b[16][mxn];
int sg[16][mxn],cnt[16][mxn];

int main()
{
	int i,j,x,y,k;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);x--;y--;
		b[x][1<<y]=1;
	}
	for(i=0;i<n;i++)
		for(j=1;j<(1<<n);j++)
			cnt[i][j]=cnt[i][j^lowbit(j)]+b[i][lowbit(j)];
	f[0]=1;
	for(i=1;i<(1<<n);i++)
	{
		if(((i&3)==3)||((i&3)==0))
		{
			f[i]=1;
			for(j=i;~j;j=(j?(j-1)&i:j-1))// mei ju zi ji
			{
				if(j==i||j==0)	continue;
				if(((j&3)!=0)&&((j&3)!=3))	continue;
				int cur=f[j^i];
				for(k=0;k<n;k++)
				{
					if(j&(1<<k))// shu yu zi ji
						cur=((ll)cur*(1ll<<cnt[k][i^j])%mdn)%mdn;
					else if(i&(1<<k))// shu yu bu ji
						cur=((ll)cur*((1ll<<cnt[k][j])-1ll)%mdn)%mdn;
				}
				f[i]=((ll)f[i]+cur)%mdn;
			}
		}
	}
	int cur=1;
	for(i=1;i<=m;i++)	cur=(cur<<1ll)%mdn;
	printf("%d\n",((ll)cur-f[(1<<n)-1]+mdn)%mdn);
	return 0;
}