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
// Zende bad Shoma nasime faghat !
#include <bits/stdc++.h>

#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " : " << x << '\n'

using namespace std;

typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;

const ll Mod = 1000000007LL;
const int N = (1 << 19) + 10;
const ll Inf = 2242545357980376863LL;
const ll Log = 30;

ll mul(ll a, ll b){
	return (a * b) % Mod;
}
ll bin_pow(ll b, ll p){
	ll res = 1;
	for(ll j = 1, pw = b; j <= p; j <<= 1, pw = mul(pw, pw))
		if(p & j)
			res = mul(res, pw);
	return res;
}
ll inv(ll x){
	return bin_pow(x, Mod - 2);
}

int n, m, m2;
int p[N];
int mk[N];
ll Sm[N];

vector< pair<int, ll> > V;
ll dp[N], iv[N];
ll Ans;

void DFS(int u){
	if(!mk[u]) return;
	if(u > m) return;
	DFS(u << 1);
	DFS(u << 1 | 1);
	Ans += mul(u, mul(Sm[u << 1], Sm[u << 1 | 1]));
	Sm[u] = mul(u, (Sm[u << 1] + Sm[u << 1 | 1]));
}

ll ans[N];

void Solve(int u){
	int L = u;
	while(L <= m) L <<= 1;
	int R = u;
	while(R <= m) R = R << 1 | 1;
	//cerr << "! " << L << ' ' << R << '\n';
	int mt;
	for(int v = L; v <= R; v++){
		mt = m + p[v];
		//if(u == 1) cerr << v << ' ' << mt << '\n';

		Sm[mt] = mul(mt, mul(dp[v], iv[u]));
		mk[mt] = 1;
		while(mt >>= 1) mk[mt] = 1;
		//cerr << "MTF\n";
	}
	//debug("P!");
	Ans = 0;
	DFS(1);
	Ans %= Mod;
	ans[u] = mul(Ans, u);
	//debug("P#");
	for(int v = L; v <= R; v++){
		mt = m + p[v];
		Sm[mt] = 0;
		mk[mt] = 0;
		while(mt >>= 1) Sm[mt] = 0, mk[mt] = 0;
	}
}

int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	dp[1] = 1;
	for(int i = 2; i < N; i++) dp[i] = mul(dp[i / 2], i);
	for(int i = 1; i < N; i++) iv[i] = inv(dp[i]);

	cin >> n;
	m = (1 << (n - 1)) - 1;
	m2 = (1 << n) - 1;
	for(int i = m + 1; i <= m2; i++) cin >> p[i];

	for(int i = m; i > 0; i--){
		//cerr << i << '\n';
		Solve(i);
	}
	ll fin = 0;
	for(int i = 1; i <= m; i++){
		ans[i] -= mul(ans[i << 1], mul(i, i << 1));
		ans[i] -= mul(ans[i << 1 | 1], mul(i, i << 1 | 1));
		ans[i] %= Mod;
		fin += ans[i];
	}
	fin %= Mod;
	fin += Mod;
	cout << fin % Mod << '\n';

	return 0;
}
