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
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif
//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
typedef long double ld;
typedef long long ll;
const int mod = (int)1e9 + 7;
int mult(int a, int b) {
    return (1LL * a * b) % mod;
}
int sum(int a, int b) {
    int s = a + b;
    if (s >= mod) s -= mod;
    return s;
}
int sub(int a, int b) {
    int s = a - b;
    if (s < 0) s += mod;
    return s;
}
int pw(int a, int b) {
    if (b == 0) return 1;
    if (b & 1) return mult(a, pw(a, b - 1));
    int res = pw(a, b / 2);
    return mult(res, res);
}
int H;
const int maxN = (1 << 18) + 10;
int prod[maxN];
int P[maxN];
int dvd[maxN];
int coef[maxN];
vector < pair < int, int > > vals[maxN];

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
//    freopen("input.txt", "r", stdin);
    cin >> H;
    int tot = (1 << H) - 1;
    for (int i = (1 << (H - 1)); i <= tot; i++) {
        cin >> P[i];
        P[i] += (1 << (H - 1)) - 1;
    }
    prod[1] = 1;
    for (int i = 2; i <= tot; i++) {
        prod[i] = mult(i, prod[i / 2]);
    }
    for (int i = 1; i <= tot; i++) {
        if (i == 1) {
            dvd[i] = 1;
        }
        else {
            dvd[i] = pw(prod[i], mod - 2);
        }
        dvd[i] = mult(dvd[i], dvd[i]);
        dvd[i] = mult(dvd[i], i);
    }
    for (int i = 1; i <= tot; i++) {
        if (i == 1) coef[i] = 1;
        else coef[i] = sub(dvd[i], dvd[i / 2]);
    }
    for (int i = (1 << (H - 1)); i <= tot; i++) {
        int g = P[i] / 2;
        while (g > 0) {
            vals[i].emplace_back(g, mult(prod[i], prod[P[i]]));
            g /= 2;
        }
        sort(vals[i].begin(), vals[i].end());
    }
    int ans = 0;
    for (int i = ((1 << (H - 1)) - 1); i >= 1; i--) {
        int c1 = 2 * i;
        int c2 = 2 * i + 1;
        int pt1 = 0;
        int pt2 = 0;
        while (pt1 < vals[c1].size() && pt2 < vals[c2].size()) {
            if (vals[c1][pt1].first < vals[c2][pt2].first) {
                vals[i].emplace_back(vals[c1][pt1]);
                pt1++;
            }
            else if (vals[c2][pt2].first < vals[c1][pt1].first) {
                vals[i].emplace_back(vals[c2][pt2]);
                pt2++;
            }
            else {
//                cout << vals[c1][pt1].first << " " << vals[c2][pt2].first << " PP  " << endl;
                assert(vals[c1][pt1].first == vals[c2][pt2].first);
                ans = sum(ans, mult(dvd[i], mult(coef[vals[c1][pt1].first], mult(vals[c1][pt1].second, vals[c2][pt2].second))));
//                cout << dvd[i] << " " << coef[vals[c1][pt1].first] << " WUT " << vals[c1][pt1].second << " " << vals[c2][pt2].second << " " << c2 << endl;
                vals[i].emplace_back(vals[c1][pt1].first, sum(vals[c2][pt2].second, vals[c1][pt1].second));
                pt1++;
                pt2++;
            }
        }
        while (pt1 < vals[c1].size()) {
            vals[i].emplace_back(vals[c1][pt1]);
            pt1++;
        }
        while (pt2 < vals[c2].size()) {
            vals[i].emplace_back(vals[c2][pt2]);
            pt2++;
        }
        vals[c1].clear();
        vals[c2].clear();
    }
    cout << ans;
    return 0;
}