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 mod = 1000000007;
int h, l, clk, ans, p[1 << 17], a[1 << 18], tag[1 << 18];
void go(int u, int prd) {
	while (u > 1) {
		prd = 1ll * prd * u % mod;
		if (tag[u] != clk) { a[u] = 0; tag[u] = clk; }
		a[u] = (a[u] + prd) % mod;
		u >>= 1;
	}
}
void clc(int u, int prd) {
	while (u > 1) {
		prd = 1ll * prd * u % mod;
		int fa = u >> 1;
		int v = u ^ 1;
		if (tag[v] == clk) {
			ans = (ans + 1ll * prd * a[v] % mod * fa) % mod;
		}
		u >>= 1;
	}
}
void dfs2(int u, int prd) {
	if (u >= l) {
		go(p[u - l] + l, prd);
		return;
	}
	dfs2(u << 1, 1ll * prd * (u << 1) % mod);
	dfs2(u << 1 | 1, 1ll * prd * (u << 1 | 1) % mod);
}
void dfs3(int u, int prd) {
	if (u >= l) {
		clc(p[u - l] + l, prd);
		return;
	}
	dfs3(u << 1, 1ll * prd * (u << 1) % mod);
	dfs3(u << 1 | 1, 1ll * prd * (u << 1 | 1) % mod);
}
void gao(int u) {
	clk++;
	dfs2(u << 1, u << 1);
	dfs3(u << 1 | 1, 1ll * u * (u << 1 | 1) % mod);
}
void dfs(int u) {
	if (u >= l) return;
	gao(u);
	dfs(u << 1);
	dfs(u << 1 | 1);
}
int main() {
	scanf("%d", &h);
	l = 1 << (h - 1);
	for (int i = 0; i < l; i++) {
		scanf("%d", &p[i]);
		p[i]--;
	}
	dfs(1);
	printf("%d\n", ans);
	return 0;
}