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 rep(i,x,y) for(int i=x,I=y+1;i<I;++i)
#define per(i,x,y) for(int i=x,I=y-1;i>I;--i)
using namespace std;
using ll=long long;
void cmax(int&x,const int&y){x=x>y?x:y;}
void cmin(int&x,const int&y){x=x<y?x:y;}
template<class T>istream&operator>>(istream&in,vector<T>&V){for(auto&x:V)in>>x;return in;}
template<class T>ostream&operator<<(ostream&out,const vector<T>&V){for(auto x:V)out<<x<<' ';return out;}
template<class T>void sort(vector<T>&V){sort(V.begin(),V.end());}
template<class T>void reverse(vector<T>&V){reverse(V.begin(),V.end());}
template<class T>int SZ(const vector<T>&V){return (int)V.size();}
void debug(){cerr<<"whxorz"<<'\n';}
int h,n;
const int N=262144;
int id[N];
const int mod=1e9+7;
int mul(const int&x,const int&y){return 1ll*x*y%mod;}
void upd(int&x,const int&y){x+=y;if(x>=mod)x-=mod;}
int tag[N];
int ans=0;
void trv2(int p,int prd,int op){
	if(p==1)return;prd=mul(prd,p);
	if(op)upd(tag[p],prd);
	else upd(ans,mul(mul(p>>1,tag[p^1]),prd));
	trv2(p>>1,prd,op);
}
int st[N],top=0;
void trv(int p,int prd,int op){prd=mul(prd,p);
	if(p>=n)return st[++top]=id[p],trv2(id[p],prd,op);
	trv(p<<1,prd,op);trv(p<<1|1,prd,op);
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cin>>h;
	n=1<<h-1;
	rep(i,1,n)cin>>id[i+n-1],id[i+n-1]+=n-1;
	//link i+n-1,id[i+n-1]
	rep(i,1,n-1){trv(i<<1,i,1);trv(i<<1|1,1,0);while(top)for(int x=st[top--];x;x>>=1)tag[x]=0;}
	cout<<ans<<'\n';
	return 0;
}