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 _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif

#include <bits/stdc++.h>

using namespace std;

const int MOD = 1000000007;

#define MUL(a,b) ((long long)(a) * (b) % MOD)

int main()
{
#ifdef _MSC_VER
	freopen("d.in", "r", stdin);
#endif
	std::ios::sync_with_stdio(false);
	int h;
	cin >> h;
	int l = (1 << (h - 1));
	vector<int> p(l);
	for (int i = 0; i < l; ++i) {
		cin >> p[i];
		--p[i];
	}

	int wc = 0;
	vector<int> c(l * 2);
	vector<int> val(l * 2);

	std::function<void(int, int)> dfs1 = [&](int id, int multi) {
		multi = MUL(multi, id);
		if (id >= l) {
			for (int next_id = p[id - l] + l; next_id > 0; next_id /= 2) {
				multi = MUL(multi, next_id);
				if (c[next_id] != wc) {
					c[next_id] = wc;
					val[next_id] = 0;
				}
				val[next_id] = (val[next_id] + multi) % MOD;
			}
			return;
		}
		dfs1(id * 2, multi);
		dfs1(id * 2 + 1, multi);
	};
	int ret = 0;
	std::function<void(int, int)> dfs2 = [&](int id, int multi) {
		multi = MUL(multi, id);
		if (id >= l) {
			for (int next_id = p[id - l] + l; next_id > 0; next_id /= 2) {
				multi = MUL(multi, next_id);
				if (next_id > 1 && c[next_id ^ 1] == wc) {
					ret = (ret + MUL(MUL(multi, val[next_id ^ 1]), next_id / 2)) % MOD;
				}
			}
			return;
		}
		dfs2(id * 2, multi);
		dfs2(id * 2 + 1, multi);
	};
	for (int r = 1; r < l; r++) {
		++wc;
		dfs1(r * 2, r);
		dfs2(r * 2 + 1, 1);
	}
	cout << ret << endl;
	return 0;
}
