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 = 300000, P = 1000000007;
int H, ans, a[maxn], b[maxn];

int main() {
    scanf("%d", &H);
    int L = 1 << (H - 1);
    for (int i = 1; i <= L; i++) {
        scanf("%d", &a[L + i - 1]), a[L + i - 1] += L - 1;
    }
    for (int i = 1; i < L; i++) {
        vector<int> V;
        function<void(int, int)> dfs1 = [&](int pos, int cur) {
            if (pos >= L) {
                for (pos = a[pos]; pos; pos >>= 1) {
                    (b[pos] += cur) %= P, V.push_back(pos);
                    cur = 1LL * cur * pos % P;
                }
                return;
            }
            dfs1(pos << 1, 1LL * cur * (pos << 1) % P);
            dfs1(pos << 1 | 1, 1LL * cur * (pos << 1 | 1) % P);
        };
        dfs1(i << 1, 1LL * i * (i << 1) % P);
        function<void(int, int)> dfs2 = [&](int pos, int cur) {
            if (pos >= L) {
                int sum = 0;
                for (pos = a[pos]; pos; pos >>= 1) {
                    cur = 1LL * cur * pos % P;
                    ans = (ans + 1LL * b[pos] * cur % P) % P;
                    ans = (ans - 1LL * pos * sum % P + P) % P;
                    sum = (1LL * sum * pos % P * pos % P + (1LL * b[pos] * cur %
                        P - 1LL * pos * sum % P + P) * pos) % P;
                }
                return;
            }
            dfs2(pos << 1, 1LL * cur * (pos << 1) % P);
            dfs2(pos << 1 | 1, 1LL * cur * (pos << 1 | 1) % P);
        };
        dfs2(i << 1 | 1, i << 1 | 1);
        for (int x : V) b[x] = 0;
    }
    printf("%d\n", ans);
    return 0;
}