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<cstdio>
#include<algorithm>
using namespace std;
int n, H;
int w[1<<18];
long long F[1<<18], Mod = 1000000007, IT[18][1<<18], D[30][30], Ans, InvF[1<<18], InvF2[1<<18];
long long Go(int a){
    if(a==1)return 1;
    return Go(a/2)*a%Mod;
}
void Put(int dep, int a, int x){
    a+=n;
    while(a){
        IT[dep][a]+=x;
        a>>=1;
    }
}
long long Pow(long long a, long long b){
    long long r=1;
    while(b){
        if(b&1)r=r*a%Mod;
        a=a*a%Mod;b>>=1;
    }
    return r;
}
void DFS(int a, int dep){
    int i, j;
    int bb=a,ee=a;
    while(bb<n){
        bb*=2,ee=ee*2+1;
    }
    bb-=n,ee-=n;
    for(i=bb;i<=ee;i++){
        Put(dep, w[i], F[n+i]*F[n+w[i]]%Mod);
    }

    if(dep!=H-1){
        DFS(a*2,dep+1);
        DFS(a*2+1,dep+1);
    }
    else{
        for(i=0;i<H;i++){
            int x = n+w[a-n];
            for(j=H-1;j>=0;j--){
                D[i][j]=IT[i][x];
                x>>=1;
            }
        }
        long long tp = F[a]*F[n+w[a-n]]%Mod;
        long long ss=0;
        for(i=0;i<H-1;i++){
            long long s=0;
            for(j=0 ;j<H-1;j++){
                D[i][j]-=D[i+1][j]+D[i][j+1]-D[i+1][j+1];
                if(D[i][j]){
                    int x = n+w[a-n];
                    s=(s+D[i][j]%Mod*InvF2[x>>(H-j-1)])%Mod;
                }
            }
            ss=(ss + s*InvF2[a>>(H-i-1)])%Mod;
        }
        Ans=(Ans+ss*tp)%Mod;
    }

    for(i=bb;i<=ee;i++){
        Put(dep, w[i], -(F[n+i]*F[n+w[i]]%Mod));
    }
}
int main(){
    int i;
    scanf("%d",&H);
    n = 1<<(H-1);
    for(i=0;i<n;i++){
        scanf("%d",&w[i]);
        w[i]--;
    }
    F[0]=1;
    for(i=1;i<n+n;i++){
        F[i]=Go(i);
    }
    for(i=0;i<n+n;i++)InvF[i]=Pow(F[i],Mod-2);
    for(i=0;i<n+n;i++)InvF2[i]=InvF[i]*InvF[i/2]%Mod;
    DFS(1,0);
    Ans=(Ans+Mod)%Mod;
    Ans=Ans*Pow(2,Mod-2)%Mod;
    printf("%lld\n",Ans);
}