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>
using namespace std;
#define ri register int
typedef long long ll;
const int N=20,M=N*N,mod=1e9+7,S=1<<15|1;
int f[S],g[S][N],power[M];
int main(){
	ri n,m,i,u,s,t,x,y,res;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)
		scanf("%d%d",&x,&y),
		--x,--y,
		g[1<<y][x]=1;
	u=(1<<n)-1;
	for(s=1;s<=u;++s)
		for(i=0;i<n;++i)
			g[s][i]=g[s&-s][i]+g[s^s&-s][i];
	power[0]=1;
	for(i=1;i<=max(n,m);++i) power[i]=power[i-1]*2ll%mod;
	f[0]=1;
	for(s=0;s<=u;++s)
		if(f[s]){
		//	printf("f %d %d\n",s,f[s]);
			for(t=s+1|s;t<=u;t=t+1|s){
				x=t^s;
				if(x>>0&1^x>>1&1)
					continue;
				res=1;
				for(i=0;i<n;++i){
					if(s&1<<i)
						res=(ll)res*power[g[x][i]]%mod;
					else if(~t&1<<i)
						res=(ll)res*(power[g[x][i]]-1)%mod;
				}
			//	printf("upd %d %d %d %d\n",s,t,x,res);
				f[t]=(f[t]+(ll)f[s]*res)%mod;
			}
		}
	printf("%d\n",(power[m]-f[u]+mod)%mod);
	return 0;
}