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
#include <bits/stdc++.h>
using namespace std;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline int rd() {
	char ch=nc(); int i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}

const int N=17, mod=1e9+7;
inline int add(int x,int y) {return (x+y>=mod) ? (x+y-mod) : (x+y);}
inline int mul(int x,int y) {return (long long)x*y%mod;}
inline int pct(int x) {return __builtin_popcount(x);}
int n,m,g[N],bin[N],f[1<<N];
inline int valid(int s1,int al) {
	if((s1&1) && (s1&2)) return 0;
	int s2=(bin[n]-1)^al, sum=1;
	for(int j=0;j<n;j++) if(s2&bin[j]) {
		if(!(s1&g[j])) return 0;
		int c=pct(s1&g[j]);
		sum=mul(sum,bin[c]-1);
	} 
	for(int j=0;j<n;j++) if(s1&bin[j]) {
		int c=pct(s2&g[j]);
		sum=mul(sum,bin[c]);
	}
	return sum;
}
int main() {
	n=rd(), m=rd();
	for(int i=0;i<=n;i++) bin[i]=1<<i;
	for(int i=1;i<=m;i++) {
		int x=rd()-1, y=rd()-1;
		g[x]|=bin[y];
	}
	f[0]=1;
	for(int s=1;s<bin[n];++s) {
		for(int s2=s;s2;s2=(s2-1)&s) {
			int t=valid(s2,s);
			if(t) f[s]=add(f[s],mul(f[s^s2],t));
		}
	}
	cout<<f[bin[n]-1]<<'\n';
}