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>
#define bitcnt(x) __builtin_popcount(x)
#define mem(a,b) memset(a,b,sizeof(a))
#define  For(i,a,b) for(int i=a,i##E=b;i<=i##E;++i)
#define rFor(i,a,b) for(int i=a,i##E=b;i>=i##E;--i)
using namespace std;
typedef long long LL;
const int N=16;
const int M=300;
const LL mod=1e9+7;
template<typename T>inline bool chkmin(T &a,const T &b){return a>b?a=b,1:0;}
template<typename T>inline bool chkmax(T &a,const T &b){return a<b?a=b,1:0;}
template<typename T>inline void read(T &x)
{
	x=0;int _f(0);char ch=getchar();
	while(!isdigit(ch))_f|=(ch=='-'),ch=getchar();
	while( isdigit(ch))x=x*10+ch-'0',ch=getchar();
	x=_f?-x:x;
}
inline void file()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
#endif
}
int n,m;
int w[N];
LL P2[M],dp[1<<N];
int main()
{
	//file();
	int x,y;
	read(n),read(m);
	P2[0]=1;
	For(i,1,m)
	{
		P2[i]=P2[i-1]*2%mod;
		read(x),read(y);
		w[x]|=(1<<(y-1));
	}
	dp[0]=1;
	int tot=(1<<n)-1;
	For(x,1,tot)
		for(int s=x;;s=(s-1)&x)
		{
			int ss=x^s,flag=(s&1?1:0)^(s&2?1:0);
			if(flag)continue;
			LL ret1=0,ret2=1;
			For(i,1,n)
			{
				if(ss&(1<<(i-1)))ret1+=bitcnt(w[i]&s);
				if(s &(1<<(i-1)))ret2=ret2*(P2[bitcnt(w[i]&ss)]-1)%mod;
			}
			dp[x]=(dp[x]+dp[s]*ret2%mod*P2[ret1]%mod)%mod;
			if(s==0)break;
		}
	printf("%lld\n",(P2[m]+mod-dp[tot])%mod);
	return 0;
}
