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<bits/stdc++.h>
#define mod 1000000007
using namespace std;
int dp[1<<15],n,m,c[15];
int qpow(int a,int b){
	int ans=1,tmp=a;
	for(;b;b>>=1,tmp=1ll*tmp*tmp%mod)
		if(b&1)ans=1ll*ans*tmp%mod;
	return ans;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0,u,v;i<m;++i)
		scanf("%d%d",&u,&v),c[u-1]|=1<<v-1;
	dp[0]=1;
	for(int i=1;i<(1<<n);++i)if((i&3)==3||(i&3)==0){
		for(int j=i;;j=(j-1)&i)if((j&3)==3||(j&3)==0){
			int k=i^j,l=0,z=dp[j];
			for(int p=0;p<n;++p)if((1<<p)&k)
				l+=__builtin_popcount(j&c[p]);
			z=1ll*z*qpow(2,l)%mod;
			
			for(int p=0;p<n;++p)if((1<<p)&j)
				z=1ll*z*(qpow(2,__builtin_popcount(k&c[p]))+mod-1)%mod;
		//	printf("<%d,%d>",j,z);
			dp[i]=(dp[i]+z)%mod;
			if(!j)break;
		}
	//	printf("[%d:%d]",i,dp[i]);
	}
	printf("%d",(qpow(2,m)-dp[(1<<n)-1]+mod)%mod);
}