Problem:
Score : 1000 points

Problem Statement

Inspired by the tv series Stranger Things, bear Limak is going for a walk between two mirror worlds.

There are two perfect binary trees of height H,
each with the standard numeration of vertices from 1 to 2^H-1.
The root is 1 and the children of x are 2 \cdot x and 2 \cdot x + 1.

Let L denote the number of leaves in a single tree, L = 2^{H-1}.

You are given a permutation P_1, P_2, \ldots, P_L of numbers 1 through L.
It describes L special edges that connect leaves of the two trees.
There is a special edge between vertex L+i-1 in the first tree and vertex L+P_i-1 in the second tree.

drawing for the first sample test, permutation P = (2, 3, 1, 4), special edges in green

Let's define product of a cycle as the product of numbers in its vertices.
Compute the sum of products of all simple cycles that have exactly two special edges, modulo (10^9+7).

A simple cycle is a cycle of length at least 3, without repeated vertices or edges.

Constraints

2 \leq H \leq 18

1 \leq P_i \leq L where L = 2^{H-1}

P_i \neq P_j (so this is a permutation)

Input

Input is given from Standard Input in the following format (where L = 2^{H-1}).

H
P_1 P_2 \cdots P_L

Output

Compute the sum of products of simple cycles that have exactly two special edges.
Print the answer modulo (10^9+7).

Sample Input 1

3
2 3 1 4

Sample Output 1

121788

The following drawings show two valid cycles (but there are more!) with products
2 \cdot 5 \cdot 6 \cdot 3 \cdot 1 \cdot 2 \cdot 5 \cdot 4 = 7200
and 1 \cdot 3 \cdot 7 \cdot 7 \cdot 3 \cdot 1 \cdot 2 \cdot 5 \cdot 4 \cdot 2 = 35280.
The third cycle is invalid because it doesn't have exactly two special edges.

Sample Input 2

2
1 2

Sample Output 2

36

The only simple cycle in the graph indeed has two special edges, and the product of vertices is
1 \cdot 3 \cdot 3 \cdot 1 \cdot 2 \cdot 2 = 36.

Sample Input 3

5
6 14 15 7 12 16 5 4 11 9 3 10 8 2 13 1

Sample Output 3

10199246

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<cstdio>
#define RI register int
#define CI const int&
using namespace std;
const int N=1<<18,mod=1e9+7;
int h,n,p[N],tag[N],ans,stk[N],top;
inline void travel2(CI x,int prod,CI opt)
{
	if (x==1) return; prod=1LL*prod*x%mod;
	if (opt) (tag[x]+=prod)%=mod; else (ans+=1LL*(x>>1)*tag[x^1]%mod*prod%mod)%=mod;
	travel2(x>>1,prod,opt);
}
inline void travel1(CI x,int prod,CI opt)
{
	prod=1LL*prod*x%mod;
	if (x>=n) return stk[++top]=n+p[x-n]-1,travel2(n+p[x-n]-1,prod,opt);
	travel1(x<<1,prod,opt); travel1(x<<1|1,prod,opt);
}
int main()
{
	RI i; for (scanf("%d",&h),n=1<<h-1,i=0;i<n;++i)
	scanf("%d",&p[i]); for (i=1;i<n;++i)
	{
		travel1(i<<1,i,1); travel1(i<<1|1,1,0);
		while (top) for (int x=stk[top--];x;x>>=1) tag[x]=0;
	}
	return printf("%d",ans),0;
}