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>

template <class T>
inline void read(T &res)
{
	res = 0; bool bo = 0; char c;
	while (((c = getchar()) < '0' || c > '9') && c != '-');
	if (c == '-') bo = 1; else res = c - 48;
	while ((c = getchar()) >= '0' && c <= '9')
		res = (res << 3) + (res << 1) + (c - 48);
	if (bo) res = ~res + 1;
}

const int N = 270000, djq = 1e9 + 7;

int h, n, p[N], f[N], inv[N], a[N], ai[N], ans;

void init(int p)
{
	if (p >= n) return;
	a[p] = 1ll * a[p >> 1] * p % djq;
	ai[p] = 1ll * ai[p >> 1] * inv[p] % djq;
	init(p << 1); init(p << 1 | 1);
}

void jjd(int u, int w)
{
	if (u >= (1 << h - 1))
	{
		for (int v = p[u]; v; v >>= 1) f[v] = (1ll * w * a[p[u]]
			% djq * a[u] + f[v]) % djq;
		return;
	}
	jjd(u << 1, w); jjd(u << 1 | 1, w);
}

void jdj(int u, int w)
{
	if (u >= (1 << h - 1))
	{
		for (int v = p[u]; v > 1; v >>= 1)
			ans = (1ll * (f[v >> 1] - f[v] + djq) * ai[v >> 2] % djq * ai[v >> 1]
				% djq * a[p[u]] % djq * w % djq * a[u] + ans) % djq;
		return;
	}
	jdj(u << 1, w); jdj(u << 1 | 1, w);
}

void djj(int u)
{
	if (u >= (1 << h - 1))
	{
		for (int v = p[u]; v; v >>= 1) f[v] = 0;
		return;
	}
	djj(u << 1); djj(u << 1 | 1);
}

int main()
{
	read(h); n = 1 << h; inv[1] = a[0] = ai[0] = 1;
	for (int i = 2; i < n; i++) inv[i] = 1ll * (djq - djq / i) * inv[djq % i] % djq;
	for (int i = 1 << h - 1; i < n; i++) read(p[i]), p[i] += 1 << h - 1, p[i]--;
	init(1);
	for (int i = 1; i < (1 << h - 1); i++)
		jjd(i << 1, ai[i >> 1]), jdj(i << 1 | 1, ai[i]), djj(i << 1);
	return std::cout << ans << std::endl, 0;
}