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>
#define F first
#define S second
#define pb push_back
using namespace std;
const int maxn=434567;
const int M=1000000007;
typedef long long ll;
typedef pair<int,ll> pi;
void add(ll &x,ll y){x+=y;if(x>=M)x-=M;}
int n,h,p[maxn];
vector<pi> v[2];
ll ans,cnt[2][maxn];
void ins(int x,ll val,int o){
    while (x){
        val=val*x%M;
        add(cnt[o][x],val);
        x>>=1;
    }
}
ll dfs2(int x){
    if (!cnt[0][x]||!cnt[1][x]) return 0;
    ll ret=(cnt[0][x*2]*cnt[1][x*2+1]+cnt[1][x*2]*cnt[0][x*2+1])%M*x%M;
    add(ret,dfs2(x*2));
    add(ret,dfs2(x*2+1));
    return ret;
}
void dfs(int x,ll val,int o){
    val=val*x%M;
    if (x>=n){
        v[o].pb((pi){p[x-n]+n,val});
        return;
    }
    dfs(x*2,val,o);
    dfs(x*2+1,val,o);
}
ll solve(int x){
    v[0].clear(); v[1].clear();
    dfs(x*2,1,0);
    dfs(x*2+1,1,1);
    for (auto x:v[0]) ins(x.F,x.S,0);
    for (auto x:v[1]) ins(x.F,x.S,1);
    ll ret=dfs2(1);
    for (auto x:v[0]) ins(x.F,M-x.S,0);
    for (auto x:v[1]) ins(x.F,M-x.S,1);
    return ret;
}
int main(){
    cin >> h; n=1<<(h-1);
    for (int i=0;i<n;i++){
        cin >> p[i];
        p[i]--;
    }
    for (int i=1;i<n;i++) ans=(ans+solve(i)*i)%M;
    cout << ans << endl;
    return 0;
}
