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;
int const p=1e9+7;
int c[1000005],w[1000005],lg2[1000005];
int mod(int x){return x>=p?x-p:x;}
int main()
{
    int h,ans=0;
    scanf("%d",&h);
    for(int i=(1<<(h-1));i<(1<<h);i++)scanf("%d",&c[i]),c[i]+=(1<<(h-1))-1;
    for(int i=2;i<(1<<(h-1));i++)lg2[i]=lg2[i>>1]+1;
    for(int i=1;i<(1<<(h-1));i++)
    {
        int t=lg2[i]+1;
        for(int j=0;j<(1<<(h-t-1));j++)
        {
            int x=(i<<(h-t))+j,now=x,prod=i;
            while(now!=i)prod=1ll*prod*now%p,now>>=1;
            now=c[x];
            while(now)prod=1ll*prod*now%p,w[now]=mod(w[now]+prod),now>>=1;
        }
        for(int j=(1<<(h-t-1));j<(1<<(h-t));j++)
        {
            int x=(i<<(h-t))+j,now=x,prod=1,pre=0;
            while(now!=i)prod=1ll*prod*now%p,now>>=1;
            now=c[x];
            while(now)
            {
                ans=mod(ans+1ll*(w[now]-1ll*w[pre]*now%p+p)*prod%p);
                prod=1ll*prod*now%p,pre=now,now>>=1;
            }
        }
        for(int j=0;j<(1<<(h-t-1));j++)
        {
            int now=c[(i<<(h-t))+j];
            while(now)w[now]=0,now>>=1;
        }
    }
    printf("%d",ans);
    return 0;
}