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<stdio.h>
#include<vector>
#define it register int
#define ct const int
#define il inline
using namespace std;
typedef long long ll;
#define rll register ll
#define cll const ll
#define P 1000000007
const int N=1000005;
int h,n,ans,f[N],p[N];
vector<int> vec;
il void mo(int &p,ct q){p+=q,p=(p>=P?p-P:p);}
il void DFS(it x,it now){
	if(x>=n){
		x=p[x-n+1]+n-1,now=(0ll+now)*x%P;
		while(x) vec.push_back(x),mo(f[x],now),x>>=1,now=(0ll+now)*x%P;
		return;
	}
	DFS(x<<1,(0ll+now)*(x<<1)%P),DFS(x<<1|1,(0ll+now)*(x<<1|1)%P);
}
il void dfs(it x,it now){
	if(x>=n){
		x=p[x-n+1]+n-1,now=(0ll+now)*x%P;it fa;
		for(fa=x,x>>=1;x;fa=x,x>>=1)
			ans=(ans+(0ll+now)*f[x])%P,ans=(ans-(0ll+now)*f[fa]%P*x)%P,ans=(ans<0?ans+P:ans),now=(0ll+now)*x%P;
		return;
	}
	dfs(x<<1,(0ll+now)*(x<<1)%P),dfs(x<<1|1,(0ll+now)*(x<<1|1)%P);
}
il void solve(ct x){
	DFS(x<<1|1,x<<1|1),dfs(x<<1,(0ll+x)*(x<<1)%P);
	for(const int &j : vec) f[j]=0;vec.clear();
}
int main(){ 
	scanf("%d",&h),n=1<<h-1;it i;
	for(i=1;i<=n;++i) scanf("%d",&p[i]);
	for(i=1;i<n;++i) solve(i);
	printf("%d",ans);
	return 0;
}