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
/**Lucky Boy**/
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define FORD(i, a, b) for (int i = (a); i >= (b); --i)
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define maxn 270000
#define maxm 1000006
#define maxc 1000000007
#define pii pair <int,int>
#define Task ""
template <typename T> inline void read(T &x){char c;bool nega=0;while((!isdigit(c=getchar()))&&(c!='-'));if(c=='-'){nega=1;c=getchar();}x=c-48;while(isdigit(c=getchar())) x=x*10+c-48;if(nega) x=-x;}
template <typename T> inline void writep(T x){if(x>9) writep(x/10);putchar(x%10+48);}
template <typename T> inline void write(T x){if(x<0){putchar('-');x=-x;}writep(x);putchar(' ');}
template <typename T> inline void writeln(T x){write(x);putchar('\n');}
using namespace std;
const int mod = maxc;
int Mul(int a,int b) {return 1ll*a*b%mod;}
int Add(int a,int b) {a += b;if (a >= mod) a -= mod;return a;}
int n,a[maxn],ans,sum[maxn];
vector <int> de;
void Dfs2(int u,int cur,bool left)
{
    if (u == 1) return;
    cur = Mul(cur,u);
    if (left)
    {
        sum[u] = Add(sum[u],cur);
    }
    else
    {
        ans = Add(ans,Mul(u >> 1,Mul(sum[u ^ 1],cur)));
    }
    Dfs2(u >> 1,cur,left);
}
void Dfs1(int u,int cur,bool left)
{
    cur = Mul(cur,u);
    if (u >= n)
    {
        int v = a[u];
        Dfs2(v,cur,left);
        de.pb(v);
        return;
    }
    Dfs1(2*u,cur,left);
    Dfs1(2*u+1,cur,left);
}
int main()
{
    srand(time(NULL));
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie();
    if(fopen(".inp", "r")) freopen(".inp", "r", stdin);
    //freopen(Task".inp", "r",stdin);
    //freopen(Task".out", "w",stdout);
    read(n);
    //cin >> n;
    n = 1 << (n-1);
    FOR(i,1,n)
    {
        //cin >> a[i + n - 1];
        read(a[i+n-1]);
        a[i + n - 1] += n - 1;
    }
    FOR(i,1,n - 1)
    {
        Dfs1(2*i,i,1);
        Dfs1(2*i+1,1,0);
        for (int &u : de)
        {
            while (u)
            {
                sum[u] = 0;
                u >>= 1;
            }
        }
        de.clear();
    }
    write(ans);
    //cout << ans;
    return 0;
}
