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;

const int K = 18, N = 1<<K, M = 1e9+7;
int prod[N][K];
int a[N];

vector<pair<int, int>> vp[N];

void add(int x, int z, int mul) {
    assert(vp[x].empty() || vp[x].back().first <= z);
    if (vp[x].empty() || vp[x].back().first < z)      vp[x].push_back({z, mul});
    else {
        vp[x].back().second += mul;
        if (vp[x].back().second >= M) vp[x].back().second -= M;
    }
}

int rgt[N];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int k;
    cin>>k;

    int L = 1<<(k-1);

    for (int i=L; i<2*L; i++) {
        prod[i][0] = i;
        for (int j=1; j<k; j++) {
            prod[i][j] = (1LL*prod[i][j-1]*(i>>j))%M;
        }
    }

    for (int i=0; i<N; i++)   vp[i].reserve(K);

    for (int i=L; i<2*L; i++) {
        int x;
        cin>>x;
        a[i] = x+L-1;
    }

     for (int k1=k-1; k1>=0; k1--) {
        for (int k2=k-1; k2>=0; k2--) {
            for (int i=L; i<2*L; i++) {
                int x = a[i];
                int pr = (1LL*prod[i][k1]*prod[x][k2])%M;

                int a1 = (i>>k1), a2 = (x>>k2);
                add(a2, a1, pr);
            }
        }
    }

    long long ans = 0;
    for (int i=1; i<=L; i++) {
        int ll = 2*i, rr = 2*i+1;

        for (auto pr: vp[rr])       rgt[pr.first] = pr.second;

        for (auto pr: vp[ll]) {
            int x = pr.first;
            if (x == 1) continue;
            unsigned long long con = (1LL*pr.second*rgt[x^1])%M;
            con = (con*i*(x>>1))%M;
            ans += con;
        }

        for (auto pr: vp[rr])       rgt[pr.first] = 0;
    }
    cout<<ans%M<<endl;
}
