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;
#define IL inline
#define RG register
#define gi geti<int>()
#define gl geti<ll>()
#define gc getchar()
#define File(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
template<typename T>IL bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
template<typename T>IL bool chkmin(T &x,const T &y){return x>y?x=y,1:0;}
template<typename T>
IL T geti()
{
	RG T xi=0;
	RG char ch=gc;
	bool f=0;
	while(!isdigit(ch))ch=='-'?f=1:f,ch=gc;
	while(isdigit(ch))xi=xi*10+ch-48,ch=gc;
	return f?-xi:xi;
}
template<typename T>
IL void pi(T k,char ch=0)
{
	if(k<0)k=-k,putchar('-');
	if(k>=10)pi(k/10);
	putchar(k%10+'0');
	if(ch)putchar(ch);
}
/*
IL unsigned int LOG2(unsigned int x)
{
	unsigned int ret;
	__asm__ __volatile__ ("bsrl %1, %%eax":"=a"(ret):"m"(x));
	return ret;
}
*/
#define count __builtin_popcount
#define lg2(x) (31-__builtin_clz(x))
#define lowbit(x) ((x)&(-(x)))
const int N=15,M=1<<N,P=1e9+7;
int f[M],g[N],pw[M];
int main(void)
{
	#ifndef ONLINE_JUDGE
//	File("");
	#endif
	pw[0]=f[0]=1;
	int n=gi,m=gi;
	for(int i=1;i<=m;++i)pw[i]=(pw[i-1]<<1)%P;
	for(int i=1;i<=m;++i){
		int u=gi-1,v=gi-1;
		g[u]|=1<<v;
	}
	for(int s=1;s<(1<<n);++s)//枚举总点集
		if((s&1)==(s>>1&1))
		for(int x=s;x;x=(x-1)&s)
			if((x&1)==(x>>1&1))
			{
				int y=s^x;
				ll ans=1;
				for(int ts=s;ts;ts^=lowbit(ts))
				{
					int val=lowbit(ts),i=lg2(val);
					if(x&val)(ans*=pw[count(g[i]&y)])%=P;
					else (ans*=pw[count(g[i]&x)]-1)%=P;
				}
				(f[s]+=f[y]*ans%P)%=P;
			}
	pi((pw[m]-f[(1<<n)-1]+P)%P);
	return 0;
}