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<map> 
#include<ctime>
#include<algorithm>
#include<cstdio>
#include<set>
#include<memory.h>
#include<queue>
#include<cmath>
using namespace std;
#define ll long long
#define ld long double
#define For(i,x,y)  for(ll i=x;i<=y;++i)
#define FOr(i,x,y)  for(ll i=x;i>=y;--i)
inline ll read(){   ll 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 void write(ll x){    if (x<0) putchar('-'),x=-x; if (x>=10) write(x/10);   putchar(x%10+'0');  } 
inline void writeln(ll x){ write(x);   puts("");   }
const ll mod=1e9+7,N=(1<<10)|111;
ll f[(1<<15)|111],g[20][(1<<15)|111],zyy[(1<<15)|111],n,m,mp[20][20],all,ans,ct[(1<<15)|111];
void add(ll &x,ll y){	x=(x+y)%mod;	}
ll qang(ll s,ll t){
	ll ans=1;
	for(;zyy[s&-s];s-=s&-s){
		ll to=zyy[s&-s];
		ans=ans*((1<<g[to][t])-1)%mod;
	}return ans;
}
ll sui(ll s,ll t){
	ll ans=1;
	for(;zyy[s&-s];s-=s&-s){
		ll to=zyy[s&-s];
		ans=(ans<<g[to][t])%mod;
	}return ans;
}
int main(){
	n=read();	m=read();
	For(i,1,m){	ll x=read(),y=read();	mp[x][y]=1;	}all=(1<<n)-1;
	For(s,0,all)	For(i,1,n)	For(j,1,n)if (!(s>>(i-1)&1)&&(s>>(j-1)&1))
	g[i][s]+=mp[i][j];
	For(s,0,all)	For(i,1,n)	if (s>>(i-1)&1)	zyy[s]=i;
	ans=1;
	For(i,1,m)	add(ans,ans);
	For(s,1,all)	ct[s]=ct[s-(s&-s)]+1;
	For(s,1,all)	if (!((s>>0&1)&&(s>>1&1)))f[s]=qang(all^s,s)*sui(s,all^s)%mod;
	For(s,1,all)	for(ll t=(s-1)&s;t;t=(t-1)&s)	if (!((s>>0&1)&&(s>>1&1)&&!(t>>0&1)&&!(t>>1&1)))	add(f[s],qang(all^s,s^t)*sui(s^t,all^s)%mod*f[t]);
	writeln(f[all]%mod);
}
/*
3 3
1 2
1 3
2 3
*/