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>

#define ls(u) ((u) << 1)
#define rs(u) ((u) << 1 | 1)
#define fa(u) ((u) >> 1)

typedef long long int64;

const int N = 1 << 18, MOD = 1e9 + 7;

int n, m, h, p[N + 5], pro[N + 5], iPro[N + 5], tag[N + 5];

inline void getLeaf(int u, std::vector<int> &vec) {
	int len = 32 - __builtin_clz(u);
	for (int i = 0; i < (1 << (h - len)); i++) {
		vec.emplace_back((u << (h - len)) + i);
	}
}
inline int solve(int u) {
	std::vector<int> l, r;
	getLeaf(ls(u), l);
	getLeaf(rs(u), r);
	for (auto v : l) {
		int w = (int64)pro[v] * iPro[fa(u)] % MOD;
		for (v = p[v]; v > 0; v = fa(v)) {
			w = (int64)w * v % MOD;
			tag[v] = (tag[v] + w) % MOD;
		}
	}
	int ans = 0;
	for (auto v : r) {
		int w = (int64)pro[v] * iPro[u] % MOD;
		for (v = p[v]; v > 1; v = fa(v)) {
			w = (int64)w * v % MOD;
			ans = (ans + (int64)tag[v ^ 1] * w % MOD * fa(v) % MOD) % MOD;
		}
	}
	for (auto v : l) {
		for (v = p[v]; v > 0; tag[v] = 0, v = fa(v));
	}
	return ans;
}
int main() {
	scanf("%d", &h);
	n = 1 << h, m = n >> 1;
	for (int i = m; i < n; i++) {
		scanf("%d", &p[i]);
		p[i] += m - 1;
	}
	pro[0] = iPro[0] = iPro[1] = 1;
	for (int i = 2; i < n; i++) {
		iPro[i] = (int64)(MOD - MOD / i) * iPro[MOD % i] % MOD;
	}
	for (int i = 1; i < n; i++) {
		pro[i] = (int64)pro[fa(i)] * i % MOD;
		iPro[i] = (int64)iPro[fa(i)] * iPro[i] % MOD;
	}
	int ans = 0;
	for (int i = 1; i < m; i++) {
		ans = (ans + solve(i)) % MOD;
	}
	printf("%d\n", ans);
	return 0;
}