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;
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define lc 2 * v
#define rc 2 * v + 1
#define mid (s + e) / 2
#define ll long long
#define int long long
#define ld long double
#define pii pair <int , int>
#define pll pair <long long , long long>
#define FAST ios::sync_with_stdio(false);cin.tie(0);
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")

const ll maxn = 3e5 + 100 , N = 2e6 + 10 , SQ = 320 , base = 97 , mod = 1e9 + 7 , INF = 1e18 , lg = 19;

int H , S , E , ans , p[maxn] , pos[maxn] , sum[maxn];
vector <int> leaf;
bool mark[maxn];

inline void get(int v) {
    int s = v , e = v;
    for (int i = 0; i < H - 1; ++i) {
        s = (s * 2);
        e = (e * 2) + 1;
    }
    S = s; E = e;
    for (int i = s; i <= e; ++i) {
        leaf.PB(i);
        mark[i] = 1;
        pos[i] = i - s;
    }
}



inline void undo_root(int v) {
    while(v >= 1) {
        sum[v] = 0;
        v /= 2;
    }
}
inline void undo(int v) {
    int s = v , e = v;
    while(!mark[s] && !mark[e]) {
        s = (2 * s);
        e = (2 * e + 1);
    }
    for (int i = s; i <= e; ++i) {
        int idx = pos[i];
        undo_root(leaf[p[idx]]);
    }
}


inline void go_root1(int v , int val) {
    while(v >= 1) {
        val = (val * v) % mod;
        sum[v] += val;
        v /= 2;
    }
}
inline void go_root2(int v , int val) {
    while(v > 1) {
        val = (val * v) % mod;
        int par = v / 2;
        int other_child = (par * 2 == v ? v + 1 : v - 1);
        ans = (ans + (((par * sum[other_child]) % mod) * val) % mod) % mod;
        v /= 2;
    }
}


void go_l(int v , int val) {
    val = (val * v) % mod;
    if(mark[v]) {
        int idx = pos[v];
        go_root1(leaf[p[idx]] , val);
        return;
    }
    go_l(2 * v , val);
    go_l(2 * v + 1 , val);
}
void go_r(int v , int val) {
    val = (val * v) % mod;
    if(mark[v]) {
        int idx = pos[v];
        // cout << leaf[idx] << ' ' << leaf[p[idx]] << endl;
        go_root2(leaf[p[idx]] , val);
        return;
    }
    go_r(2 * v , val);
    go_r(2 * v + 1 , val);
}



void fix_lca(int v) {
    go_l(2 * v , v);
    go_r(2 * v + 1 , 1);
    undo(2 * v);
}

int32_t main() {
	FAST
    cin >> H;
    for (int i = 0; i < (1 << (H - 1)); ++i) {
        cin >> p[i]; p[i]--;
    }
    get(1);

    for (int i = 1; i < leaf[0]; ++i) {
        fix_lca(i);
    }
    cout << ans << '\n';
}
