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 <iostream>
#include <cstdio>
#define MN 262144
#define mod 1000000007
using namespace std;
int H, p[MN + 5], ans, root[MN + 5], cnt;
struct Tree {
  int l, r, val, tag;
}T[MN * 18 + 5];

inline void Mark(int x, int v) {
  if (!T[x].tag) T[x].tag = 1;
  T[x].val = 1LL * T[x].val * v % mod;
  T[x].tag = 1LL * T[x].tag * v % mod;
}

inline void pushdown(int x) {
  Mark(T[x].l, T[x].tag);
  Mark(T[x].r, T[x].tag);
  T[x].tag = 0;
}

int Build(int x, int b, int val, int multiplier) {
  int k = ++cnt; 
  if (x == val) {
    T[k].val = 1LL * x * multiplier % mod;
    return k;
  }
  if (val & (1 << b)) {
    T[k].r = Build(x << 1 | 1, b - 1, val, multiplier);
    T[k].val = 1LL * T[T[k].r].val * x % mod;
  } else {
    T[k].l = Build(x << 1, b - 1, val, multiplier);
    T[k].val = 1LL * T[T[k].l].val * x % mod;
  }
  return k;
}

int Merge(int x, int y, int k, int multiplier) {
  //cout << "Merge" << x << " " << y << " " << k << endl;
  if (!x || !y) {
    int r = x + y;
    Mark(r, multiplier);
    return r;
  }
  if (T[x].tag) pushdown(x);
  if (T[y].tag) pushdown(y);
  ans = (ans + (1LL * T[T[x].l].val * T[T[y].r].val + 1LL * T[T[x].r].val * T[T[y].l].val) % mod * k % mod * multiplier) % mod;
  //cout << "MMMMM" << (1LL * T[T[x].l].val * T[T[y].r].val + 1LL * T[T[x].r].val * T[T[y].l].val) % mod * k % mod * multiplier % mod << endl;
  T[x].l = Merge(T[x].l, T[y].l, k << 1, multiplier);
  T[x].r = Merge(T[x].r, T[y].r, k << 1 | 1, multiplier);
  T[x].val = 1LL * (T[x].val + T[y].val) * multiplier % mod;
  return x;
}

void Dfs(int x) {
  if ((x << 1) >= (1 << H)) {
    root[x] = Build(1, H - 2, p[x], x);
    //cout << "root[" << x << "]=" << root[x] << endl;
    return;
  }
  Dfs(x << 1);
  Dfs(x << 1 | 1);
  root[x] = Merge(root[x << 1], root[x << 1 | 1], 1, x);
}

int main() {
  scanf("%d", &H);
  for (int i = 1 << H - 1; i < 1 << H; ++i) {
    scanf("%d", &p[i]);
    p[i] += (1 << H - 1) - 1;
  }
  Dfs(1);
  cout << ans;
  return 0;
}