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;
using ll = long long;
using pll = pair <ll, ll>;

const ll mod = 1e9 + 7;

ll P[303030], D[303030];
ll n, h, ans;

void f(ll p, ll s, ll e)
{
	if(s == e)return;

	vector <ll> V;
	ll m, i, x, y, t, v;

	m = s + e >> 1;
	f(p << 1, s, m);
	f(p << 1 | 1, m + 1, e);

	for(i = s; i <= m; i ++){
		for(x = i + n - 1, t = 1; x != p; x >>= 1){
			t = t * x % mod;
		}
		for(y = P[i] + n - 1; y; y >>= 1){
			t = t * y % mod;
			D[y] += t; if(D[y] >= mod) D[y] -= mod;
			V.push_back(y);
		}
	}

	for(i = m + 1; i <= e; i ++){
		for(x = i + n - 1, t = p; x != p; x >>= 1){
			t = t * x % mod;
		}
		for(y = P[i] + n - 1, v = 0; y; y >>= 1){
			v = v * y % mod;
			ans = (ans + (D[y] - v) * t) % mod;
			v = D[y]; t = t * y % mod;
		}
	}

	for(ll &t: V) D[t] = 0;
}

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);

	ll i;

	cin >> h; n = 1 << h - 1;

	for(i = 1; i <= n; i ++){
		cin >> P[i];
	}

	f(1, 1, n);

	cout << (ans + mod) % mod << "\n";

	return 0;
}