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<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<cstring>
#include<climits>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
inline int read() {
	bool f=0;int x=0;char c=getchar();
	while(c<'0'||'9'<c){if(c==EOF)exit(0);if(c=='-')f=1;c=getchar();}
	while('0'<=c&&c<='9') x=(x<<3)+(x<<1)+(c^48),c=getchar();
	return !f?x:-x;
}
#define MAXN 15
#define INF 0x3f3f3f3f
#define Mod (int)(1e9+7)
int G[MAXN+5];
LL f[(1<<MAXN)+5];
int bitcnt(int x){
	int ret=0;
	while(x)
		ret++,x-=(x&(-x));
	return ret;
}
int main(){
	int n=read(),m=read();
	for(int i=1;i<=m;i++){
		int u=read()-1,v=read()-1;
		G[u]|=(1<<v);
	}
	f[0]=1;
	for(int U=0;U<=(1<<n)-1;U++)
		if((U&1)==((U>>1)&1)){
			for(int S=U;S;S=(S-1)&U)
				if((S&1)==((S>>1)&1)){
					int T=U^S;
					LL cnt=1;
					for(int i=0;i<n;i++)
						if((U>>i)&1){
							if((S>>i)&1)
								cnt=(1ll<<bitcnt(G[i]&T))*cnt%Mod;
							else cnt=((1ll<<bitcnt(G[i]&S))-1)*cnt%Mod;
						}
					f[U]=(f[U]+cnt*f[T]%Mod)%Mod;
				}
		}
	LL sum=1;//
	for(int i=1;i<=m;i++)
		sum=(sum<<1)%Mod;
	printf("%lld\n",(sum-f[(1<<n)-1]+Mod)%Mod);
	return 0;
}