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 <cmath>
#include <functional>
#include <fstream>
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <list>
#include <time.h>
#include <math.h>
#include <random>
#include <deque>
#include <queue>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <chrono>
#include <cstring>

using namespace std;

typedef long long ll;

#ifdef iq
  mt19937 rnd(228);
#else
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
#endif

const int M = 1e9 + 7;

int mul(int a, int b) {
  return (a * (ll) b) % M;
}

int pw(int a, int n) {
  int ret = 1;
  while (n) {
    if (n % 2 == 0) {
      a = mul(a, a);
      n /= 2;
    } else {
      ret = mul(ret, a);
      n--;
    }
  }
  return ret;
}

int inv(int x) {
  return pw(x, M - 2);
}

void add(int &a, int b) {
  a += b;
  if (a >= M) a -= M;
}

const int N = (1 << 18) + 220;

vector <pair <int, int> > f[N][2];
int sum[N][2];

int main() {
#ifdef iq
  freopen("a.in", "r", stdin);
#endif
  ios::sync_with_stdio(0);
  cin.tie(0);
  int h = 18;
  cin >> h;
  int n = (1 << h) - 1;
  vector <int> par(n + 1);
  vector <int> x(n + 1);
  vector <int> ix(n + 1);
  int l = (1 << (h - 1));
  x[1] = 1;
  for (int v = 1; v <= n; v++) {
    if (2 * v <= n) {
      x[2 * v] = mul(x[v], 2 * v);
    }
    if (2 * v + 1 <= n) {
      x[2 * v + 1] = mul(x[v], 2 * v + 1);
    }
  }
  for (int v = 1; v <= n; v++) {
    int t = inv(x[v]);
    ix[v] = mul(mul(t, t), v);
  }
  vector <int> p(l);
  for (int i = 0; i < l; i++) {
    cin >> p[i];
    p[i]--;
  }
  auto gen_path = [&] (int v) {
    vector <pair <int, int> > path;
    int x = v / 2;
    int was = v;
    while (true) {
      path.push_back({x, was % 2});
      was = x;
      if (x == 1) break;
      x /= 2;
    }
    return path;
  };
  for (int i = 0; i < l; i++) {
    int a = l + i;
    int b = l + p[i];
    auto A = gen_path(a);
    for (auto c : A) {
      f[c.first][c.second].push_back({b, x[a]});
      /*
      for (auto t : B) {
        ll id = c.first * (ll) (n + 1) + t.first;
        add(f[c.second][t.second][id], mul(x[a], x[b]));
      }
      */
    }
  }
  int ans = 0;
  for (int i= 1; i <= n; i++) {
    for (auto c : f[i][0]) {
      int v = c.first, val = c.second;
      int was = v;
      int cur = v / 2;
      while (true) {
        add(sum[cur][was % 2], mul(x[v], val));
        if (cur == 1) break;
        was = cur;
        cur /= 2;
      }
    }
    int s = 0;
    for (auto c : f[i][1]) {
      int v = c.first, val = c.second;
      int was = v;
      int cur = v / 2;
      while (true) {
        add(s, mul(ix[cur], mul(sum[cur][(was % 2) ^ 1], mul(x[v], val))));
        if (cur == 1) break;
        was = cur;
        cur /= 2;
      }
    }
    for (auto c : f[i][0]) {
      int v = c.first;
      while (true) {
        sum[v][0] = sum[v][1] = 0;
        if (v == 1) break;
        v /= 2;
      }
    }
    add(ans, mul(s, ix[i]));
  }
  cout << ans << '\n';
}
