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 fi first
#define se second

const int N = (1 << 18);
const int mod = 1e9 + 7;

using namespace std;

void add(int &x, int y)
{
        x += y;
        if(x >= mod){
                x -= mod;
        } else if(x < 0){
                x += mod;
        }
}

int H, L;
int p[N];
int A[N], a[N][18], b[N][18], f[N];
int c1[N], c2[N];
int res;

void solve(int X, int Y)
{
        int bx = (1 << X), by = (1 << Y);
        for(int i = 0, j = 0; i < L; i += bx, j += 2){
                for(int h = 0; h < bx / 2; h++){
                        c1[i + h] = j;
                }
                for(int h = bx / 2; h < bx; h++){
                        c1[i + h] = j + 1;
                }
        }
        for(int i = 0, j = 0; i < L; i += by, j += 2){
                for(int h = 0; h < by / 2; h++){
                        c2[i + h] = j;
                }
                for(int h = by / 2; h < by; h++){
                        c2[i + h] = j + 1;
                }
        }

        for(int i = 0; i < L; i++){
                A[i] = 1;
                if(c1[i] & 1){
                        A[i] = 1ll * A[i] * b[i][X] % mod;
                } else{
                        A[i] = 1ll * A[i] * a[i][X] % mod;
                }
                if(c2[p[i]] & 1){
                        A[i] = 1ll * A[i] * b[p[i]][Y] % mod;
                } else{
                        A[i] = 1ll * A[i] * a[p[i]][Y] % mod;
                }
        }

        for(int i = 0; i < L; i += bx){
                for(int j = 0; j < bx / 2; j++){
                        add(f[c2[p[i + j]]], A[i + j]);
                }
                for(int j = bx / 2; j < bx; j++){
                        add(res, 1ll * A[i + j] * f[c2[p[i + j]] ^ 1] % mod);
                }
                for(int j = 0; j < bx / 2; j++){
                        add(f[c2[p[i + j]]], - A[i + j]);
                }
        }
}

int main()
{
        ios_base::sync_with_stdio(0);

        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);

        cin >> H;
        L = (1 << H - 1);
        for(int i = 0; i < L; i++){
                cin >> p[i];
                p[i] -= 1;
        }

        for(int i = 1; i < H; i++){
                for(int j = 0, x; j < L; j++){
                        x = L + j;
                        a[j][i] = 1;
                        for(int h = 0; h < i; h++){
                                a[j][i] = 1ll * a[j][i] * x % mod;
                                x /= 2;
                        }
                        b[j][i] = 1ll * a[j][i] * x % mod;
                }
        }

        for(int i = 1; i < H; i++){
                for(int j = 1; j < H; j++){
                        solve(i, j);
                }
        }

        cout << res << "\n";
}
