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;
const int maxn=(1<<18)+233,mod=1000000007;
inline void Add(int &a,int b){a=a+b>=mod?a+b-mod:a+b;}
int H,p[maxn],sum[maxn],ans=0;
inline void addit(int x,int opt){for(;x;x>>=1)opt=1ll*opt*x%mod,Add(sum[x],opt);}
inline void Addit(int x,int opt)
{
	opt=1ll*opt*x%mod;
	if(x>=(1<<(H-1))) return (void)addit(p[x],opt);
	Addit(x<<1,opt);Addit(x<<1|1,opt);
}
inline void getans(int x,int opt){for(;x>1;x>>=1)opt=1ll*opt*x%mod,Add(ans,1ll*opt*(x>>1)%mod*sum[x^1]%mod);}
inline void Getans(int x,int opt)
{
	opt=1ll*opt*x%mod;
	if(x>=(1<<(H-1))) return (void)getans(p[x],opt);
	Getans(x<<1,opt);Getans(x<<1|1,opt);
}
int main()
{
	scanf("%d",&H);
	for(int i=(1<<(H-1));i<(1<<H);i++) scanf("%d",p+i),p[i]+=(1<<(H-1))-1;
	for(int i=1;i<(1<<(H-1));i++) Addit(i<<1,i),Getans(i<<1|1,1),Addit(i<<1,mod-i);
	cout<<ans<<endl;
}