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 Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
#define Reg register
#define RI Reg int
#define Con const
#define CI Con int&
#define I inline
#define W while
#define N (1<<18)
#define X 1000000007
#define LT rt<<1,d+1
#define RT rt<<1|1,d+1
#define Inc(x,y) ((x+=(y))>=X&&(x-=X))
using namespace std;
int H,L,ans,p[N+5],V[N+5];
I void Work(CI rt,CI d,RI v)
{
	if(v=1LL*v*rt%X,d==H) {RI x=p[rt-L]+L;W(x) v=1LL*v*x%X,Inc(V[x],v),x>>=1;}
	else Work(LT,v),Work(RT,v);
}
I void Calc(CI rt,CI d,RI v)
{
	if(v=1LL*v*rt%X,d==H) {RI x=p[rt-L]+L;W(x^1) v=1LL*v*x%X,ans=(1LL*v*(x>>1)%X*V[x^1]+ans)%X,x>>=1;}
	else Calc(LT,v),Calc(RT,v);
}
I void Travel(CI rt=1,CI d=1)
{
	if(d==H) return;Work(LT,rt),Calc(RT,1),Work(LT,X-rt),Travel(LT),Travel(RT);
}
int main()
{
	RI i;for(scanf("%d",&H),L=1<<H-1,i=0;i^L;++i) scanf("%d",p+i),--p[i];
	return Travel(),printf("%d\n",ans),0;
}