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 N 15
#define M 1<<15
#define mod 1000000007
int g[N],f[M],t[M];
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;
}
inline int add(int x,int y){
	return x+y>=mod?x+y-mod:x+y;
}
inline int sub(int x,int y){
	return x-y<0?x-y+mod:x-y;
}
inline int mul(int x,int y){
	return 1LL*x*y%mod;
}
inline int calc(int x){
	return __builtin_popcount(x);
}
signed main(){
	int n=read(),m=read();
	f[0]=t[0]=1;
	for (int i=1;i<=m;++i) t[i]=(t[i-1]<<1)%mod;
	for (int i=1;i<=m;++i){
		int x=read()-1,y=read()-1;g[x]|=(1<<y);
	}
	for (int s=1;s<(1<<n);++s){
		if ((s&1)==(s>>1&1)){
			for (int u=s;u;u=u-1&s){
				if ((u&1)==(u>>1&1)){
					int ans=1;
					for (int i=0;i<n;++i){
						if (s>>i&1){
							if (u>>i&1) ans=mul(ans,t[calc(g[i]&(s^u))]);
							else ans=mul(ans,t[calc(g[i]&u)]-1);
						}
					}
					f[s]=add(f[s],mul(f[s^u],ans));
				}
			}
		}
	}
	cout<<sub(t[m],f[(1<<n)-1])<<endl;
	return 0;
}