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;
typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f,oo = inf;
#define rd() read()
#define mem(x,v) memset(x,v,sizeof(x))
#define IL inline
#define RG register
#define rep(i,a,b) for(RG int i=(a);i<(b);++i)
#define Rep(i,a,b) for(RG int i=(a);i<=(b);++i)
#define Dep(i,a,b) for(RG int i=(a);i>=(b);--i)
#define pc putchar
#define gc getchar
IL ll read(){
	RG ll x=0;char f=0;RG char c=gc();
	for(;!isdigit(c);c=gc())f|=(c=='-');
	for(;isdigit(c);c=gc())x=(x<<1)+(x<<3)+(c^48);
	return f?-x:x;
}
#define lowbit(x) ((x & -x))
IL void write(ll x){if(x<0)x=-x,pc('-');if(x>=10)write(x/10);pc(x%10+'0');}
IL void writeln(ll x){write(x);puts("");}
const int Mod = 1e9+7;
int n,m,bin[205],g[20][40005],f[40005],q[40005];
int main(){
//	freopen("gemo.in","r",stdin);
//	freopen("gemo.out","w",stdout);
	n = rd(),m = rd();
	int ans = 0;
	Rep(i,1,m){
		int x = rd()-1,y = rd()-1;
		g[x][(1<<y)] = true;
	}
	rep(i,0,n){
		rep(j,1,(1<<n)){
			g[i][j] = g[i][j-lowbit(j)] + g[i][lowbit(j)];
		}
	}
	for(int S=0;S<(1<<n);++S){
		if((S&1) != (S>>1&1)) continue;
		f[S] = 1;
		for(int T=S;T;T=(T-1)&S){
			//s1 : T,s2 : S-T
			if((T&1) != (T>>1&1)) continue;
			int w = f[T];
			rep(j,0,n){
				//是sg为1的点 
				if(T>>j&1) w = 1ll * w * ((1 << g[j][S-T]) - 1) % Mod;
				else if((S-T)>>j&1)  w = 1ll * w * ((1 << g[j][T])) % Mod;
			}
			f[S] = (f[S] + w) % Mod;
		}
	}
	ll anss = 1;
	for(int i=1;i<=m;i++) anss = 1ll * anss * 2 % Mod;
	writeln((anss - f[(1<<n)-1] + Mod)%Mod);
	return 0;
}