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;

const int maxn = 1<<18+1;
int ivalue[maxn];
int value[maxn];
int P[maxn];

const int mod = 1e9+7;
int mul(int a,int b){
    long long t = 1ll*a*b;
    return t - t/mod*mod;
}
int add(int a,int b){
    a+=b;
    if(a>=mod)  a-=mod;
    return a;
}
int sub(int a,int b){
    a-=b;
    if(a<0) a+=mod;
    return a;
}

int main(){
    ios::sync_with_stdio(false);
    int H;
    cin>>H;
    for(int i=(1<<H-1);i<(1<<H);++i)
        cin>>P[i], P[i]+=(1<<H-1)-1, ivalue[i] = i;
    int ans = 0;
    for(int i=(1<<H-1)-1;i>=1;--i){
        int sum = 0;
        int L = i*2, R = i*2+1;
        int x = L, y = R, z;
        while((x*2)<(1<<H)) x*=2;
        while((y*2)<(1<<H)) y*=2;
        z = y + y - x;
        for(int j = x; j < y; ++j){
            int p = P[j], v = ivalue[j];
            while(p){
                v = mul(v,p);
                value[p] = add(value[p],v);
                p/=2;
            }
            ivalue[j] = mul(ivalue[j], i );
        }
        for(int j = y; j < z; ++j){
            int p = P[j], v = ivalue[j], t = 0;
            while(p){
                sum = add(sum,mul(sub(value[p],mul(t,p)),v));
                v = mul(v,p);
                t = value[p];
                p/=2;
            }
            ivalue[j] = mul(ivalue[j], i );
        }
        ans = add(ans,mul(i,sum));
        for(int j =x; j < y; ++j){
            int p = P[j];
            while(p){
                value[p]=0;
                p/=2;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}