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
#pragma GCC optimize("Ofast","inline")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

const int mod=1e9+7,N=1<<18;
int sum[N],ans,h,n,p[N];
vector<int> clear;

void dfs2(int x,int mul,bool islc){
	if(x==1) return ;
	mul=1ll*mul*x%mod;
	if(islc) sum[x]=(sum[x]+mul)%mod;
	else ans=(ans+1ll*(x>>1)*sum[x^1]%mod*mul%mod)%mod;
	dfs2(x>>1,mul,islc);
}

void dfs1(int x,int mul,bool islc){
	mul=1ll*mul*x%mod;
	if(x>=n){
		int y=n+p[x-n+1]-1;
		dfs2(y,mul,islc);
		if(islc) clear.push_back(y);
		return ;
	}
	dfs1(x<<1,mul,islc);
	dfs1(x<<1|1,mul,islc);
}

signed main(){
	read(h);
	n=1<<h-1;
	for(int i=1;i<=n;i++) read(p[i]);
	for(int lca1=1;lca1<n;lca1++){
		dfs1(lca1<<1,lca1,1);
		dfs1(lca1<<1|1,1,0);
		for(int leaf:clear) for(int x=leaf;x;x>>=1) sum[x]=0;
		clear.clear();
	}
	write(ans);
}