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;
#define ll long long
#define f(i, x, n) for(int i = x; i < (int)(n); ++i)

int const N = 1 << 18, md = 1e9 + 7;
int p[N], fr[N], H, L, an;

void ad(int &x, int y) { if ((x += y) >= md)x -= md; }

vector<pair<int, int> > go(int v = 1, int h = 1){
	if (h == H)return vector<pair<int, int> > { { v, v } };
	auto a = go(v << 1, h + 1), b = go(v << 1 | 1, h + 1);
	
	f(i, 0, a.size()){
		int u = p[a[i].first - L + 1] + L - 1, c = a[i].second;
		while (u > 1){
			c = (ll)c * u % md;
			ad(fr[u], c);
			u >>= 1;
		}
	}
	
	f(i, 0, b.size()){
		int u = p[b[i].first - L + 1] + L - 1, c = b[i].second;
		while (u > 1){
			c = (ll)c * u % md;
			ad(an, (ll)fr[u ^ 1] * c % md * (u >> 1) % md * v % md);
			u >>= 1;
		}
	}
	
	f(i, 0, a.size()){
		int u = p[a[i].first - L + 1] + L - 1;
		while (u > 1){
			fr[u] = 0;
			u >>= 1;
		}
	}
	a.insert(a.end(), b.begin(), b.end());
	b.clear();
	for (auto &t: a)t.second = (ll)t.second * v % md;
	return a;
}

int main(){
	scanf("%d", &H);
	L = 1 << H - 1;
	f(i, 1, L + 1)scanf("%d", p + i);
	go();
	printf("%d\n", an);
}