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<cstring>
using namespace std;
#define N 400020
const int mod=1e9+7;
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,p[N],a[N],cnt[N],ans;
void dfs(int u,int l,int r){
	if(l==r)return;
	int mid=(l+r)>>1;
	dfs(u<<1,l,mid),dfs(u<<1|1,mid+1,r);
	for(int i=l;i<=mid;++i){
		a[i]=1LL*a[i]*u%mod;
	}
	for(int i=l;i<=mid;++i){
		int v=(1<<n-1)+p[i]-1;
		int num=a[i];
		do{
			num=1LL*num*v%mod;
			cnt[v]=(cnt[v]+num)%mod;
			v>>=1;
		}while(v);
	} 
	int las=ans;
	for(int i=mid+1;i<=r;++i){
		int v=(1<<n-1)+p[i]-1;
		int num=a[i];
		int las=0;
		do{
			las=1LL*las*v%mod;
			ans=(ans+1LL*num*(cnt[v]-las))%mod;
			las=cnt[v];
			num=1LL*num*v%mod;
			v>>=1;
		}while(v);
	}
	for(int i=l;i<=mid;++i){
		int v=(1<<n-1)+p[i]-1;
		do{
			cnt[v]=0;
			v>>=1;
		}while(v);
	} 
	for(int i=mid+1;i<=r;++i){
		a[i]=1LL*a[i]*u%mod;
	}
}
int main(){
	n=read();
	for(int i=1;i<=(1<<n-1);++i){
		p[i]=read();
		a[i]=(1<<n-1)+i-1;
	}
	dfs(1,1,1<<n-1);
	printf("%d\n",(ans+mod)%mod);
	return 0;
}

