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 <cstdio>
#include <cstring>
using namespace std;

#define M 1000000007
#define N 0x100000
#define ll long long

#define lc(x) ((x) << 1)
#define rc(x) ((x) << 1 | 1)
#define fa(x) ((x) >> 1)

int h, p[N], l[N], r[N];
int flag[N];
int dep[N];

int main() {
	//freopen("A.in", "r", stdin);
	int h, n;
	scanf("%d", &h);
	n = 1 << (h - 1);
	for (int i = n; i < (n << 1); i++) {
		scanf("%d", p + i);
		p[i] += n - 1;
	}
	for (int i = 0; i < h; i++)
		for (int j = (1 << i); j < (2 << i); j++)
			dep[j] = i;
	for (int i = 1; i < (n << 1); i++) {
		l[i] = i << (h - dep[i] - 1);
		r[i] = l[i] + (1 << (h - dep[i] - 1)) - 1;
	}
	
	ll ans = 0;
	for (int i = 1; i < n; i++) {
		for (int j = l[lc(i)]; j <= r[lc(i)]; j++) {
			ll prod = i;
			for (int k = j; k != i; k = fa(k))
				prod = prod * k % M;
			for (int k = p[j]; k; k = fa(k)) {
				prod = prod * k % M;
				flag[k] = (flag[k] + prod) % M;
			}
		}
		for (int j = l[rc(i)]; j <= r[rc(i)]; j++) {
			ll prod = 1, last = 0;
			for (int k = j; k != i; k = fa(k))
				prod = prod * k % M;
			for (int k = p[j]; k; k = fa(k)) {
				ans += prod * (M + flag[k] - last * k % M) % M;
				prod = prod * k % M;
				last = flag[k];
			}
			ans %= M;
		}
		for (int j = l[lc(i)]; j <= r[lc(i)]; j++)
			for (int k = p[j]; k; k = fa(k))
				flag[k] = 0;
	}
	printf("%lld\n", ans);
}