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
#ifdef LC
#include "pch.h"
#else
#include <bits/stdc++.h>
#endif
using namespace std;
using ll = long long;
#define int ll
#define all(x) x.begin(), x.end()
#define x first
#define y second
#define mt make_tuple

const int H = 1e7;
int *ptr[H];
int  val[H];
int siz = 0;
void save(int &x) {
    ptr[siz] =  &x;
    val[siz++] = x;
}
void backup(int to) {
    assert(0 <= to && to <= siz);
    while (siz > to) {
        --siz;
        *ptr[siz] = val[siz];
    }
}

const int N = 1 << 20;
const int M = 1e9 + 7;

int h, n;
int p[N];
int dp[N];

int twin(int v) {
    assert(n <= v && v < 2 * n);
    v = p[v - n] + n;
    assert(n <= v && v < 2 * n);
    return v;
}
int ls(int v) {
    return 2 * v;
}
int rs(int v) {
    return 2 * v + 1;
}
int pa(int v) {
    return v / 2;
}
int br(int v) {
    return v ^ 1;
}

void dfs_add(int v, int m) {
    (m *= v) %= M;
    if (v < n) {
        dfs_add(ls(v), m);
        dfs_add(rs(v), m);
    } else {
        for (v = twin(v); v > 0; v = pa(v)) {
            (m *= v) %= M;
            save(dp[v]);
            (dp[v] += m) %= M;
        }
    }
}

int ans = 0;

void dfs_ans(int v, int m) {
    (m *= v) %= M;
    if (v < n) {
        dfs_ans(ls(v), m);
        dfs_ans(rs(v), m);
    } else {
        for ((m *= (v = twin(v))) %= M; pa(v) > 0; v = pa(v)) {
            (m *= pa(v)) %= M;
            (ans += m * dp[br(v)]) %= M;
        }
    }
}

int32_t main() {
#ifdef LC
    assert(freopen("input.txt", "r", stdin));
#endif
    ios::sync_with_stdio(0); cin.tie(0);
    
    cin >> h;
    n = 1 << (h - 1);
    for (int i = 0; i < n; ++i) {
        cin >> p[i];
        --p[i];
    }
    for (int i = 1; i < n; ++i) {
        int save = siz;
        dfs_add(ls(i), i);
        dfs_ans(rs(i), 1);
        backup(save);
    }
    cout << ans << "\n";
    return 0;
}
