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 a[300000];
int TOTOT = 32333;
int vis[300000], T[300000];
int n;

void modify(int x, long long mul) {
	x >>= 1, mul = mul * x % mod;
	while (x) {
		if (vis[x] != TOTOT) vis[x] = TOTOT, T[x] = 0;
		T[x] = (T[x] + mul) % mod;
		x >>= 1, mul = mul * x % mod;
	}
}

long long query(int x, long long mul) {
	mul = mul * x % mod, x >>= 1;
	long long lst = 0, ans = 0;
	while (x) {
		lst = lst * x % mod;
		if (vis[x] != TOTOT) vis[x] = TOTOT, T[x] = 0;
		ans += (T[x] - lst + mod) * mul % mod;
		lst = T[x], mul = mul * x % mod, x >>= 1;
	}
	return ans;
}

void modify_son(int x, long long mul) {
	if (x >= (1 << n - 1)) {
		int y = a[x - (1 << n - 1) + 1] + (1 << n - 1) - 1;
		modify(y, mul * y % mod);
		return ;
	}
	modify_son(x << 1, mul * (x << 1) % mod), modify_son(x << 1 | 1, mul * (x << 1 | 1) % mod);
}

long long query_son(int x, long long mul) {
	if (x >= (1 << n - 1)) {
		int y = a[x - (1 << n - 1) + 1] + (1 << n - 1) - 1;
		return query(y, mul * x % mod);
	}
	return (query_son(x << 1, mul * x % mod) + query_son(x << 1 | 1, mul * x % mod)) % mod;
}

int main () {
	scanf("%d", &n);
	for (int i = 1; i <= (1 << n - 1); i++) {
		scanf("%d", &a[i]);
	}
	int ans = 0;
	for (int i = 1; i < (1 << n - 1); i++) {
		TOTOT++;
		modify_son(i << 1, i << 1);
		ans = (ans + query_son(i << 1 | 1, i)) % mod;
	}
	printf("%lld\n", ans);
	return 0;
}
