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 = 1 << 20;
const int P = 1e9 + 7;
typedef long long ll;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
int n, ans, t, l[MAXN], r[MAXN], a[MAXN], b[MAXN], s[MAXN];
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
void updans(int pos, int val) {
	val = 1ll * val * pos % P;
	update(s[pos], val);
	if (pos != 1) updans(pos / 2, val);
}
void getans(int pos, int val) {
	val = 1ll * val * pos % P;
	if (pos != 1) {
		update(ans, 1ll * pos / 2 * val % P * s[pos ^ 1] % P);
		getans(pos / 2, val);
	}
}
void work(int pos, int depth) {
	if (depth == n) {
		l[pos] = r[pos] = b[pos] = pos;
		return;
	}
	work(pos * 2, depth + 1);
	work(pos * 2 + 1, depth + 1);
	l[pos] = l[pos * 2], r[pos] = r[pos * 2 + 1];
	for (int i = l[pos * 2]; i <= r[pos * 2]; i++)
		updans(a[i], 1ll * b[i] * pos % P);
	for (int i = l[pos * 2 + 1]; i <= r[pos * 2 + 1]; i++)
		getans(a[i], b[i]);
	for (int i = l[pos * 2]; i <= r[pos * 2]; i++)
		updans(a[i], P - 1ll * b[i] * pos % P);
	for (int i = l[pos]; i <= r[pos]; i++)
		b[i] = 1ll * b[i] * pos % P;
}
int main() {
	read(n), t = 1 << (n - 1);
	for (int i = t; i < t * 2; i++)
		read(a[i]), a[i] += t - 1;
	work(1, 1);
	cout << ans << endl;
	return 0;
}
