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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn (1<<18)+1
#define mod 1000000007

int H,n,p[maxn];
int vis[maxn],id=0,sum[maxn];
int ans=0;
void add(int &x,int y){x=(x+y>=mod?x+y-mod:x+y);}
void go2(int x,int prod,bool left){
	if(x==1)return;
	prod=1ll*prod*x%mod;
	if(left){
		if(vis[x]<id)vis[x]=id,sum[x]=0;
		add(sum[x],prod);
	}
	else{
		if(vis[x^1]<id)vis[x^1]=id,sum[x^1]=0;
		add(ans,1ll*(x/2)*prod%mod*sum[x^1]%mod);
	}
	go2(x/2,prod,left);
}
void go(int x,int prod,bool left){
	prod=1ll*prod*x%mod;
	if(x>=n){go2(n+p[x-n+1]-1,prod,left);return;};
	go(x*2,prod,left);
	go(x*2+1,prod,left);
}

int main()
{
	scanf("%d",&H);n=1<<(H-1);
	for(int i=1;i<=n;i++)scanf("%d",&p[i]);
	for(int i=1;i<n;i++){
		id++;
		go(i*2,i,true);
		go(i*2+1,1,false);
	}
	printf("%d",ans);
}