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 ll;
#define fi first
#define se second
const int M=262144;
const ll mod=1e9+7;
int n;
vector<pair<int,ll> >v[M];
ll s[M],s2[M];
ll a[M];
bool done[M];
ll ans=0;
ll coo(ll p,ll t){
	return p*t%mod*((1+(p/2)*(t/2)%mod*p%mod*t%mod-p*(p/2)-t*(t/2))%mod)%mod;
}
void solve(ll p,ll t){
	ll cur=s[t]*s[t]-s2[t];
	cur=(cur%mod+mod)%mod;
	ll co=coo(p,t);
	co=(co+mod)%mod;
	ans=(ans+co*cur)%mod;
	done[t]=true;
}
int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	for(int i=(1<<(n-1)); i<(1<<n) ;i++){
		cin >> a[i];
		a[i]+=(1<<(n-1))-1;
		ll p=i;
		ll r=1;
		for(int j=0; j<n ;j++){
			ll q=a[i];
			ll s=1;
			for(int k=0; k<n ;k++){
				v[p].push_back({q,r*s%mod});
				s=s*q%mod;
				q/=2;
			}
			r=r*p%mod;
			p/=2;
		}
	}
	for(int i=1; i<(1<<n) ;i++){
		for(auto c:v[i]){
			s[c.fi]=(s[c.fi]+c.se)%mod;
			s2[c.fi]=(s2[c.fi]+c.se*c.se)%mod;
		}
		for(auto c:v[i]){
			if(done[c.fi]) continue;
			solve(i,c.fi);
		}
		for(auto c:v[i]){
			done[c.fi]=false;
			s[c.fi]=s2[c.fi]=0;
			//g[c.fi].clear();
		}
	}
	ll half=(mod+1)/2;
	ans=ans*half%mod;
	cout << ans << '\n';
}