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;
#define int long long 
const int mod=1e9+7,N=1<<20;
int cnt[N],a[N],p[N],n,ans;
void solve(int u,int l,int r){
	if(l==r) return (void)(a[l]=u);
	int mid=(l+r)>>1;
	solve(u<<1,l,mid);
	solve(u<<1|1,mid+1,r);
	for(int i=l;i<=mid;++i)
		a[i]=a[i]*u%mod;
	for(int i=l;i<=mid;++i){
		int v=(1<<n-1)+p[i]-1,num=a[i];
		while(v) num=num*v%mod,cnt[v]=(cnt[v]+num)%mod,v>>=1;
	}
	for(int i=mid+1;i<=r;++i){
		int v=(1<<n-1)+p[i]-1,num=a[i],las=0;
		while(v) las=las*v%mod,ans=(ans+(cnt[v]-las+mod)%mod*num%mod)%mod,num=num*v%mod,las=cnt[v],v>>=1;
	}
	for(int i=l;i<=mid;++i){
		int v=(1<<n-1)+p[i]-1;
		while(v) cnt[v]=0,v>>=1; 
	}
	for(int i=mid+1;i<=r;++i)
		a[i]=a[i]*u%mod;
}
signed main(){
	cin>>n;
	for(int i=1;i<=(1<<n-1);++i)
		cin>>p[i];
	solve(1,1,1<<n-1);
	cout<<ans;
	return 0;
}