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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

const int MOD=1000000007;

int n,m,bin[205],g[20][40005],f[40005],q[40005];

int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int lowbit(int x) {return x&(-x);}

int main()
{
	n=read();m=read();
	bin[0]=1;
	for (int i=1;i<=max(n,m);i++) bin[i]=bin[i-1]*2%MOD;
	for (int i=1;i<=m;i++)
	{
		int x=read()-1,y=read()-1;
		g[x][bin[y]]=1;
	}
	for (int i=0;i<n;i++)
		for (int j=1;j<bin[n];j++)
			g[i][j]=g[i][j-lowbit(j)]+g[i][lowbit(j)];
	for (int i=1;i<bin[n];i++)
	{
		if ((i&1)!=(i&2)/2) continue;
		f[i]=1;int h=0;
		for (int s=i;s;s=(s-1)&i) q[++h]=s;
		while (h--)
		{
			int s1=q[h+1],s2=s1^i,w=f[s1];
			if ((s1&1)!=(s1&2)/2) continue;
			for (int j=0;j<n;j++)
				if (s1&bin[j]) w=(LL)w*(bin[g[j][s2]]-1)%MOD;
				else if (s2&bin[j]) w=(LL)w*bin[g[j][s1]]%MOD;
			f[i]+=w;f[i]-=f[i]>=MOD?MOD:0;
		}
	}
	printf("%d",(bin[m]+MOD-f[bin[n]-1])%MOD);
	return 0;
}