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;

#define ll long long
#define all(aaa) aaa.begin(), aaa.end()

const int N = 5e5 + 5, MOD = 1e9 + 7;
int p[N];
ll prod[N], sum[N];
int h;

void add(ll x, ll &y) {
    x += y;
    if (x >= MOD)
        x -= MOD;
}

ll ans = 0;

void solve(int v, int lvl = 1) {
    if (lvl == h) {
        prod[v] = v;
    }
    else {
        solve(v * 2, lvl + 1);
        solve(v * 2 + 1, lvl + 1);

        int ct = (1 << (h - lvl - 1)),
            start_left = v * 2 * ct,
            start_right = (v * 2 + 1) * ct;

        for (int i = 0; i < ct; i++) {
            int x = p[start_left + i - (1 << (h - 1))] - 1 + (1 << (h - 1));

            ll cur_prod = prod[start_left + i] * v % MOD;

            while (x > 1) {
                cur_prod = cur_prod * x % MOD;
                sum[x] = (sum[x] + cur_prod * (x / 2)) % MOD;
                x /= 2;
            }
        }

        // ll gaga = 0;

        for (int i = 0; i < ct; i++) {
            int x = p[start_right + i - (1 << (h - 1))] - 1 + (1 << (h - 1));

            ll cur_prod = prod[start_right + i];

            while (x > 1) {
                cur_prod = cur_prod * x % MOD;
                ans = (ans + sum[x ^ 1] * cur_prod) % MOD;
                // gaga = (gaga + sum[x ^ 1] * cur_prod) % MOD;
                x /= 2;
            }
        }

        // cout << v << " " << gaga << "\n";

        for (int i = 0; i < ct; i++) {
            int x = p[start_left + i - (1 << (h - 1))] - 1 + (1 << (h - 1));

            while (x > 1) {
                sum[x] = 0;
                x /= 2;
            }
        }

        for (int i = 0; i < ct * 2; i++) {
            prod[start_left + i] = prod[start_left + i] * v % MOD;
        }
    }
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

    
    cin >> h;

    int n = (1 << h);

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

    solve(1);

    cout << ans << "\n";

	return 0;
}