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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <assert.h>
#include <fstream>
#include <unordered_map>

using namespace std;

typedef long long ll;

const int MAXN = 262150;
const ll MOD = 1000000007;

int n;
int p[MAXN];

ll ans;
ll val1[MAXN];
ll val2[MAXN];
ll sum[MAXN];

pair<int,ll> mdf[MAXN][18];
pair<int,ll> qry[MAXN][18];

int main()
{
	scanf("%d",&n);
	for (int i = 0;i < (1 << (n - 1));i++)
	{
		scanf("%d",&p[i]);
		p[i]--;
	}
	for (int L = 0;L < n - 1;L++)
	{
		for (int i = 0;i < (1 << (n - 1));i++)
		{
			int u = i + (1 << (n - 1)),v = p[i] + (1 << (n - 1));
			ll vu = 1,vv = 1;
			for (int j = u;j >= 1;j >>= 1)
			{
				(vu *= j) %= MOD;
				val1[j] = vu;
			}
			for (int j = v;j >= 1;j >>= 1)
			{
				(vv *= j) %= MOD;
				val2[j] = vv;
			}
			for (int j = u,l = 0;j >= 1;j >>= 1,l++)
				if (l == L)
					for (int k = v,r = 0;k >= 1;k >>= 1,r++)
						mdf[u][r] = make_pair(k,val1[j] * val2[k] % MOD);
		}
		for (int i = 0;i < (1 << (n - 1));i++)
		{
			int u = i + (1 << (n - 1)),v = p[i] + (1 << (n - 1));
			ll vu = 1,vv = 1;
			for (int j = u;j >= 1;j >>= 1)
			{
				(vu *= j) %= MOD;
				val1[j] = vu;
			}
			for (int j = v;j >= 1;j >>= 1)
			{
				(vv *= j) %= MOD;
				val2[j] = vv;
			}
			for (int j = u,l = 0;j >= 2;j >>= 1,l++)
				if (l == L)
					for (int k = v,r = 0;k >= 2;k >>= 1,r++)
						qry[u ^ (1 << L)][r] = make_pair(k ^ 1,val1[j >> 1] * val2[k >> 1] % MOD);
		}
		for (int i = (1 << (n - 1 - L));i < (1 << (n - L));i++)
		{
			for (int j = i << L;j < (i << L) + (1 << L);j++)
				for (int k = 0;k < n;k++)
					(sum[mdf[j][k].first] += mdf[j][k].second) %= MOD;
			for (int j = i << L;j < (i << L) + (1 << L);j++)
				for (int k = 0;k < n;k++)
					(ans += sum[qry[j][k].first] * qry[j][k].second) %= MOD;
			for (int j = i << L;j < (i << L) + (1 << L);j++)
				for (int k = 0;k < n;k++)
					sum[mdf[j][k].first] = 0;
		}
	}
	printf("%lld\n",ans * ((MOD + 1) / 2) % MOD);
	return 0;
}