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>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define IT iterator
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
#define debug puts("wzpakking")
#define y1 ysghysgsygsh
using namespace std;
const int mo=1000000007;
const int N=1<<19|5;
int n,p[N],w[N],s[N],ans;
void insert(int p,int v){
	for (;p!=1;p>>=1)
		v=1ll*v*p%mo,s[p]=(s[p]+v)%mo;
}
void update(int p,int v){
	for (;p!=1;p>>=1){
		v=1ll*v*p%mo;
		ans=(ans+1ll*v*s[p^1]%mo*(p>>1))%mo;
	}
}
void solve(int k,int l,int r){
	if (l!=r){
		int mid=(l+r)/2;
		solve(k*2,l,mid);
		solve(k*2+1,mid+1,r);
		For(i,l,mid) insert(p[i]+(1<<(n-1))-1,w[i]);
		For(i,mid+1,r) update(p[i]+(1<<(n-1))-1,1ll*w[i]*k%mo);
		For(i,l,mid) insert(p[i]+(1<<(n-1))-1,mo-w[i]);
	}
	For(i,l,r)
		w[i]=1ll*w[i]*k%mo;
}
int main(){
	scanf("%d",&n);
	For(i,0,(1<<(n-1))-1) scanf("%d",&p[i]);
	For(i,0,(1<<(n-1))-1) w[i]=1;
	solve(1,0,(1<<(n-1))-1);
	cout<<ans<<endl;
}