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>
#define reg register
#define ll long long
#define inf 0x3f3f3f3f3f3f3f3fll
#define next liuyifan_next
using namespace std;
const ll p=1e9+7;
ll n,m,g[10000005],f[10000005],pow2[10000005],x,y,ans;
inline ll add(reg ll x,reg ll y) 
{
    return (x+=y)>p?x-p:x;
}
int main()
{
    cin>>n>>m;
    f[0]=pow2[0]=1; 
    for(reg ll i=1;i<=m;i++)pow2[i]=(pow2[i-1]<<1)%p; 
    for(reg ll i=1;i<=m;i++)
	{
		cin>>x>>y;
		x--;
		y--;
		g[x]|=(1<<y);
	} 
    for(reg ll s=1;s<1<<n;s++)
	if((s&1)==(s>>1&1))
    for(reg ll u=s;u;u=u-1&s) 
	if((u&1)==(u>>1&1)) 
    {
        ans=1;
        for(reg ll i=0;i<n;i++) 
		if(s>>i&1) 
        {
            if(u>>i&1)ans=ans*pow2[__builtin_popcount(g[i]&(s^u))]%p;
            else ans=ans*(pow2[__builtin_popcount(g[i]&u)]-1)%p; 
        }
        f[s]=add(f[s],f[s^u]*ans%p); 
    }
    cout<<(pow2[m]-f[(1<<n)-1]+p)%p<<endl;
    return 0;
}