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<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=18,M=262144,mod=1000000007;

int add(int a,int b,int p=mod){return a+b>=p?a+b-p:a+b;}
int sub(int a,int b,int p=mod){return a-b<0?a-b+p:a-b;}
int mul(int a,int b,int p=mod){return (LL)a*b%p;}
void sadd(int &a,int b,int p=mod){a=add(a,b,p);}
void ssub(int &a,int b,int p=mod){a=sub(a,b,p);}
void smul(int &a,int b,int p=mod){a=mul(a,b,p);}

int n,a[M+9];

void into(){
  scanf("%d",&n);
  for (int i=1<<n-1;i<1<<n;++i){
	scanf("%d",&a[i]);
    a[i]+=(1<<n-1)-1;
  }
}

int now[M+9];

void Dfs_now0(int k,int v){
  if (k>=1<<n-1){
	for (smul(v,k=a[k]);k;smul(v,k>>=1)) sadd(now[k],v);
    return;
  }
  Dfs_now0(k<<1,mul(v,k<<1));
  Dfs_now0(k<<1|1,mul(v,k<<1|1));
}

int Dfs_ans(int k,int v){
  if (k>=1<<n-1){
	int res=0,t=0;
	for (smul(v,k=a[k]);k;smul(v,k>>=1)) sadd(res,mul(now[k^1],mul(v,k>>1)));
	return res;
  }
  return add(Dfs_ans(k<<1,mul(v,k<<1)),Dfs_ans(k<<1|1,mul(v,k<<1|1)));
}

void Dfs_now1(int k){
  if (k>=1<<n-1){
	for (k=a[k];k;k>>=1) now[k]=0;
    return;
  }
  Dfs_now1(k<<1);
  Dfs_now1(k<<1|1);
}

int ans;

void Get_ans(){
  for (int i=1;i<1<<n-1;++i){
	Dfs_now0(i<<1,mul(i,i<<1));
	sadd(ans,Dfs_ans(i<<1|1,i<<1|1));
	Dfs_now1(i<<1);
  }
}

void work(){
  Get_ans();
}

void outo(){
  printf("%d\n",ans);
}

int main(){
  into();
  work();
  outo();
  return 0;
}