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;

typedef pair <int, int> ii;
typedef long long ll;

const int Maxh = 18;
const int mod = 1000000007;

int H, n;
int L;
int P[1 << Maxh];
int sum[1 << Maxh][2];
vector <ii> has[1 << Maxh];
int res;

void Add(int ind, int val)
{
    int v = L + P[ind];
    val = ll(val) * v % mod;
    while (v > 1) {
        int u = v / 2;
        val = ll(val) * u % mod;
        sum[u][v % 2] = (sum[u][v % 2] + val) % mod;
        v = u;
    }
}

void Rem(int ind, int val)
{
    int v = L + P[ind];
    val = ll(val) * v % mod;
    while (v > 1) {
        int u = v / 2;
        val = ll(val) * u % mod;
        sum[u][v % 2] = 0;
        v = u;
    }
}

void Get(int ind, int val)
{
    int v = L + P[ind];
    val = ll(val) * v % mod;
    while (v > 1) {
        int u = v / 2;
        res = (res + ll(val) * sum[u][1 - v % 2]) % mod;
        val = ll(val) * u % mod;
        v = u;
    }
}

int main()
{
    scanf("%d", &H); n = (1 << H) - 1;
    L = 1 << (H - 1);
    for (int i = 0; i < L; i++) {
        scanf("%d", &P[i]); P[i]--;
        has[L + i].push_back(ii(L + i, i));
    }
    for (int i = L - 1; i >= 1; i--) {
        int rig = 2 * i + 1, lef = 2 * i;
        has[i].resize(has[rig].size() + has[lef].size());
        int pnt = 0;
        for (int j = 0; j < has[rig].size(); j++) {
            has[i][pnt] = ii(ll(has[rig][j].first) * i % mod, has[rig][j].second);
            Add(has[i][pnt].second, has[i][pnt].first);
            pnt++;
        }
        int lim = pnt;
        for (int j = 0; j < has[lef].size(); j++) {
            Get(has[lef][j].second, has[lef][j].first);
            has[i][pnt] = ii(ll(has[lef][j].first) * i % mod, has[lef][j].second);
            pnt++;
        }
        for (int j = 0; j < lim; j++)
            Rem(has[i][j].second, has[i][j].first);
    }
    printf("%d\n", res);
    return 0;
}
