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

int d, ans;
int p[N + 5], sum[N + 5];

#define mp make_pair

typedef pair<int, int> pii;

void modify(int u, int v, int t) {
    while (u > 0) {
        v = 1ll * v * u % mo;
        sum[u] = (sum[u] + v * t) % mo;
        u = u >> 1;
    }
}

int query(int u, int v) {
    int res = 0;
    for (int u0 = 0; u > 0; u >>= 1) {
        if (u0) {
            res = (res + 1ll * v * (sum[u] - 1ll * sum[u0] * u % mo)) % mo;
        }
        u0 = u;
        v = 1ll * v * u % mo;
    }
    return res;
}

vector<pii> dfs(int u, int dep) {
    if (dep == d) {
        return vector<pii> (1, mp(u, p[u]));
    }
    
    vector<pii> res;
    vector<pii> L = dfs(u << 1, dep + 1), R = dfs(u << 1 | 1, dep + 1);
    for (auto x : L) {
        modify(x.second, 1ll * x.first * u % mo, +1);
        res.push_back(mp(1ll * x.first * u % mo, x.second));
    }

    for (auto x : R) {
        ans = (ans + query(x.second, x.first)) % mo;
        res.push_back(mp(1ll * x.first * u % mo, x.second));
    }

    for (auto x : L) {
        modify(x.second, 1ll * x.first * u % mo, -1);
    }
    return res;
}

int main() {
#ifdef Wearry
    freopen("in", "r", stdin);
#endif

    scanf("%d", &d);
    for (int i = 1 << (d-1); i < (1 << d); ++i) {
        scanf("%d", p + i);
        p[i] += (1 << (d-1)) - 1;
    }

    dfs(1, 1);
    printf("%d\n", (ans + mo) % mo);

    return 0;
}
