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>

using namespace std;

const int P=1000000007;
const int N=15;
const int S=1<<N;
const int M=N*N;

int f[S],bitcnt[S];
int tov[N];
int pw[M];
int n,m;

void update(int &x,int y){(x+=y)%=P;}

void dp()
{
	f[0]=1;
	for (int s=1;s<1<<n;++s)
		if (!((s&1)^((s>>1)&1)))
		{
			for (int u=s,t;u;u=(u-1)&s)//u sg(x)=0,t sg(x)!=0
			{
				t=s^u;
				if (!((t&1)^((t>>1)&1)))
				{
					int tmp=1;
					for (int i=0;i<n;++i)
						if ((s>>i)&1)
							if ((t>>i)&1) tmp=1ll*tmp*((pw[bitcnt[tov[i]&u]]+P-1)%P)%P;
							else tmp=1ll*tmp*pw[bitcnt[tov[i]&t]]%P;
					update(f[s],1ll*f[t]*tmp%P);
				}
			}
		}
}

int main()
{
	//freopen("daggame.in","r",stdin),freopen("daggame.out","w",stdout);
	scanf("%d%d",&n,&m),pw[0]=1;
	for (int i=1,x,y;i<=m;pw[i]=(pw[i-1]<<1)%P,++i) scanf("%d%d",&x,&y),tov[x-1]|=1<<y-1;
	for (int s=1;s<1<<n;++s) bitcnt[s]=bitcnt[s-(s&-s)]+1;
	dp(),printf("%d\n",(pw[m]+P-f[(1<<n)-1])%P);
	//fclose(stdin),fclose(stdout);
	return 0;
}