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=1000009;
const int P=1000000007;

int H,L;
int p[maxn];

long long f[maxn];
vector<int>V;
void pre(int x,long long now){
	if(x>=L){
		x=p[x-L+1]+L-1;now=now*x%P;
		while(x){
			V.push_back(x);
			f[x]=(f[x]+now)%P;
			x>>=1;now=now*x%P;
		}
		return;
	}
	pre(x<<1,now*(x<<1)%P);
	pre(x<<1|1,now*(x<<1|1)%P);
}
long long ans;
void solve(int x,long long now){
	if(x>=L){
		x=p[x-L+1]+L-1;now=now*x%P;int y=0;
		for(y=x,x=x>>1;x;y=x,x=x>>1){
			ans=(ans+now*f[x])%P;
			ans=(ans-now*f[y]%P*x%P+P)%P;
			now=now*x%P;
		}
		return;
	}
	solve(x<<1,now*(x<<1)%P);
	solve(x<<1|1,now*(x<<1|1)%P);
}
void work(int o){
	pre(o<<1|1,o<<1|1);
	solve(o<<1,1LL*o*(o<<1)%P);
	
	for(int i=0;i<V.size();++i)f[V[i]]=0;
	V.clear();
}
int main(){
	scanf("%d",&H);L=1<<(H-1);
	for(int i=1;i<=L;++i)scanf("%d",&p[i]);
	for(int i=1;i<L;++i)work(i);
	
	printf("%lld\n",ans);
	
	return 0;
}