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 D;
const int maxn=530003,mod=1000000007;
int Plus(int x,int y){return (x+=y)>=mod?x-mod:x;}
void Pluseq(int &x,int y){if((x+=y)>=mod)x-=mod;}
int Minus(int x,int y){return Plus(x,mod-y);}
void Minuseq(int &x,int y){Pluseq(x,mod-y);}
int mul(D x,int y){return x*y%mod;}
void muleq(int &x,int y){x=mul(x,y);}
int inv[maxn],n,p[maxn],ver[maxn],tot,pre[maxn],a[maxn],preinv[maxn],dp[maxn],RES,ANS;
bool vis[maxn];
void init(int u){
	if((u<<1)>n){ver[++tot]=u;return;}
	init(u<<1),init(u<<1|1);
}
void dfs(int u){
	if(u>n)return;
	int son=vis[u<<1]+vis[u<<1|1],v=vis[u<<1]*(u<<1)+vis[u<<1|1]*(u<<1|1);
	if(!son)dp[u]=a[u];
	if(son==1)dfs(v),dp[u]=dp[v];
	if(son==2){
		dfs(u<<1),dfs(u<<1|1);
		dp[u]=Plus(dp[u<<1],dp[u<<1|1]);
		Pluseq(RES,mul(mul(mul(dp[u<<1],dp[u<<1|1]),preinv[u]),preinv[u>>1]));
	}
}
int solve(int i){
	if(i>n)return 0;
	tot=0;
	init(i);
	for(int j=1;j<=tot;j++){
		int u=p[ver[j]];
		a[u]=mul(pre[ver[j]],pre[u]);
		for(;u&&!vis[u];u>>=1)vis[u]=1;
	}
	RES=0;
	dfs(1);
	for(int j=1;j<=tot;j++){
		int u=p[ver[j]];
		for(;u&&vis[u];u>>=1)vis[u]=0;
	}
	return RES;
}
int main(){
	inv[1]=1;
	for(int i=2;i<maxn;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
	scanf("%d",&n);
	n=(1<<n)-1;
	for(int i=(n>>1)+1;i<=n;i++)scanf("%d",p+i),p[i]+=(n>>1);
	pre[0]=preinv[0]=1;
	for(int i=1;i<=n;i++)pre[i]=mul(pre[i>>1],i),preinv[i]=mul(preinv[i>>1],inv[i]);
	for(int i=1;i<=n;i++){
		Pluseq(ANS,mul(mul(Minus(solve(i),Plus(solve(i<<1),solve(i<<1|1))),preinv[i]),preinv[i>>1]));
	}
	printf("%d\n",ANS);
	return 0;
}