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;
const int N=1<<15|1,p=1000000007;
int F[N],f[15][N],k[16*16],S[16],i,j,n,m,x,X;
char c;

inline char getc()
{
	#define VV 10000000
	static char s[VV],*l=s,*r=s;
	if (l==r)
		l=s,r=s+fread(s,1,VV,stdin);
	return l==r?0:*l++;
}

int read(){ int z=0; do c=getc(); while (c<'0'||c>'9'); while (c>='0'&&c<='9') z*=10,z+=c-'0',c=getc(); return z; }

void Plus(int &a,int b){ a+=b,a-=a<p?0:p; }

bool check(int s){ return !((s&1)^(s>>1&1)); }

void solve()
{
	n=read(),m=read(),X=(1<<n)-1,k[0]=1;
	for (register int i=1;i<=m;++i) x=read()-1,S[x]|=1<<read()-1,k[i]=(k[i-1]<<1)%p;
	for (register int s=1;s<=X;++s)
	{
		for (register int i=0;i<n;++i)
			if (!(s>>i&1))
				f[i][s]+=__builtin_popcount(S[i]&s);
	}
	F[0]=1;
	for (register int s1=0;s1<X;++s1)
		if (check(s1))
			for (register int s=X^s1,s2=s;s2;s2=s2-1&s)
				if (check(s2))
				{
					register int s3=s^s2,z=F[s1];
					for (register int i=0;i<n;++i)
					{
						if (s2>>i&1)
							z=(LL)z*k[f[i][s3]]%p;
						if (s3>>i&1)
							z=(LL)z*(k[f[i][s2]]-1)%p;
					}
					if (z)
						Plus(F[s1^s2],z);
				}
	printf("%d\n",(k[m]-F[X]+p)%p);
}

int main()
{
	solve();
	return 0;
}