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 <cstdio>
#include <unordered_map>

typedef long long LL;
const int Mod = 1000000007, Inv2 = (Mod + 1) / 2;
const int MN = 1 << 18 | 7;

int H, N, P[MN];
int Inv[MN], f[MN], g[MN];

int tag[MN], Ans;

int main() {
	scanf("%d", &H), --H, N = 1 << H;
	f[0] = f[1] = g[0] = g[1] = 1;
	Inv[1] = 1;
	for (int i = 2; i < 2 * N; ++i) {
		Inv[i] = (LL)(Mod - Mod / i) * Inv[Mod % i] % Mod;
		f[i] = (LL)f[i / 2] * i % Mod;
		g[i] = (LL)g[i / 2] * Inv[i] % Mod;
	}
	for (int i = N - 1; i >= 1; --i) g[i] = (LL)g[i] * g[i / 2] % Mod;
	for (int i = N; i < 2 * N; ++i) scanf("%d", &P[i]), P[i] += N - 1;
	for (int a = 0; a < H; ++a) {
		for (int o = 0; o < 1 << a; ++o) {
			int i = o | 1 << a, Sum = 0;
			for (int q = 0; q < 1 << (H - a - 1); ++q) {
				int j = i << (H - a) | q;
				int val = (LL)f[j] * f[P[j]] % Mod;
				for (int b = 0; b < H; ++b) {
					int k = P[j] >> b;
					tag[k] = (tag[k] + val) % Mod;
				}
			}
			for (int q = 0; q < 1 << (H - a - 1); ++q) {
				int j = i << (H - a) | q | 1 << (H - a - 1);
				int t_Sum = 0;
				for (int b = 0; b < H; ++b) {
					int k = P[j] >> b ^ 1;
					t_Sum = (t_Sum + (LL)tag[k] * g[k / 2]) % Mod;
				}
				Sum = (Sum + (LL)t_Sum * f[j] % Mod * f[P[j]]) % Mod;
			}
			Ans = (Ans + (LL)Sum * g[i]) % Mod;
			for (int q = 0; q < 1 << (H - a - 1); ++q) {
				int j = i << (H - a) | q;
				for (int b = 0; b < H; ++b) tag[P[j] >> b] = 0;
			}
		}
	}
	printf("%d\n", Ans);
	return 0;
}