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.

graph for sample1

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).

Examples

Input

3
2 3 1 4


Output

121788


Input

2
1 2


Output

36


Input

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


Output

10199246
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#define maxh 20
#define maxn 200005
using namespace std;
const long long mo=1e9+7;
long long h,n,p[maxn];
long long ans=0,tag[maxn*2];
void modify(long long x,long long data)
{
	if(!x) return;
	modify(x/2,data*x%mo);
	tag[x]=(tag[x]+data)%mo;
}
void qk(long long x,long long data)
{
	if(!x) return;
	qk(x/2,data*x%mo);
	tag[x]=(tag[x]-data)%mo;
}
long long que(long long x,long long data,long long a)
{
	if(!x) return 0;
	long long res=data*(tag[x]+mo-a)%mo;
	res+=que(x/2,data*(x/2)%mo,tag[x]*x%mo);
	return res%mo;
}
void work(long long x,long long data)
{
	if(x>=n)
	{
		modify(p[x-n+1]+n-1,data);
		return;
	}
	work(x*2,data*x*2%mo);
	work(x*2+1,data*(x*2+1)%mo);
}
void clear(long long x,long long data)
{
	if(x>=n)
	{
		qk(p[x-n+1]+n-1,data);
		return;
	}
	clear(x*2,data*x*2%mo);
	clear(x*2+1,data*(x*2+1)%mo);
}
long long calc(long long x,long long data)
{
	if(x>=n)  return que(p[x-n+1]+n-1,data*(p[x-n+1]+n-1)%mo,0);
	long long res=0;
	res+=calc(x*2,data*x*2%mo);
	res+=calc(x*2+1,data*(x*2+1)%mo);
	return res%mo;
}
int main()
{
	scanf("%lld",&h);
	n=(1<<(h-1));
	for(long long i=1;i<=n;i++) scanf("%lld",&p[i]);
	for(long long i=1;i<n;i++)
	{
		//memset(tag,0,sizeof(tag)); 
		work(i*2,i*2*i%mo);
		ans=(ans+calc(i*2+1,i*2+1))%mo;
		clear(i*2,i*2*i%mo);
	}
	printf("%lld",ans);
	return 0; 
} 