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
#define DIN freopen("input.txt","r",stdin);
#define DOUT freopen("output.txt","w",stdout);
#include <bits/stdc++.h>
#include <cstdio>
#define mem(a,b) memset(a,b,sizeof(a))
#define REP(i,a,b) for(int i=(a);i<=(int)(b);i++)
#define REP_(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
using namespace std;
typedef long long LL;
typedef std::vector<int> VI;
typedef std::pair<int,int> PI;
int read()
{
    int x=0,flag=1; char c=getchar();
    while((c>'9' || c<'0') && c!='-') c=getchar();
    if(c=='-') flag=0,c=getchar();
    while(c<='9' && c>='0') {x=(x<<3)+(x<<1)+c-'0';c=getchar();}
    return flag?x:-x;
}

#define mod M
#define k (1<<n-1)
const int M=1e9+7;
const int maxn=1e6+5;
int n,p[maxn],d[maxn],ans;

void solve(int x,int y)
{
    y=1ll*y*x%M;
    if(x<(1<<n-1))
    {
        solve(x<<1,y);
        solve(x<<1|1,y);
    }
    else
    {
        int yy=y;
        for(int i=p[x];i>1;i>>=1)
        {
            yy=1ll*yy*i%M;
            int s=(d[i>>1]-1ll*d[i]*(i>>1)%M+M)%M;
            ans=(ans+1ll*s*yy)%M;
        }
    }
}

void res(int x)
{
    if(x<(1<<n-1))
    {
        res(x<<1);
        res(x<<1|1);
    }
    else
    {
        for(int i=p[x];i;i>>=1) d[i]=0;
    }
}

void cal(int x,int y)
{
    y=1ll*y*x%M;
    if(x<(1<<n-1))
    {
        cal(x<<1,y);
        cal(x<<1|1,y);
    }
    else
    {
        for(int i=p[x];i;i>>=1)
        {
            y=1ll*y*i%M;
            d[i]=(d[i]+y)%M;
        }
    }
}

int main()
{
    n=read();
    for(int i=(1<<n-1);i<(1<<n);i++)
        p[i]=read()+(1<<n-1)-1;
    for(int i=1;i<(1<<n-1);i++)
    {
        cal(i<<1,i);
        solve(i<<1|1,1);
        res(i<<1);
    }
    cout<<ans;

    return 0;
}
