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<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<iostream>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include<bitset>
#include<vector>
#include<limits.h>
#include<assert.h>
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define FOR(I, A, B) for (int I = (A); I <= (B); ++I)
#define FORS(I, S) for (int I = 0; S[I]; ++I)
#define RS(X) scanf("%s", (X))
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define CASET int ___T; scanf("%d", &___T); for(int cs=1;cs<=___T;cs++)
#define MP make_pair
#define PB emplace_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PII> VPII;
typedef pair<LL,LL> PLL;
typedef vector<PLL> VPLL;
template<class T> void _R(T &x) { cin >> x; }
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template<class T, class... U> void R(T &head, U &... tail) { _R(head); R(tail...); }
template<class T> void _W(const T &x) { cout << x; }
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template<class T,class U> void _W(const pair<T,U> &x) {_W(x.F); putchar(' '); _W(x.S);}
template<class T> void _W(const vector<T> &x) { for (auto i = x.begin(); i != x.end(); _W(*i++)) if (i != x.cbegin()) putchar(' '); }
void W() {}
template<class T, class... U> void W(const T &head, const U &... tail) { _W(head); putchar(sizeof...(tail) ? ' ' : '\n'); W(tail...); }
#ifdef HOME
 #define DEBUG(...) {printf("[DEBUG] ");W(__VA_ARGS__);}
#else
 #define DEBUG(...)
#endif
int MOD = 1e9+7;
void ADD(LL& x,LL v){x=(x+v)%MOD;if(x<0)x+=MOD;}
/*}}}*/
const int SIZE = 1<<18;
LL an;
VPII top_tree[SIZE];
int from[SIZE];
struct Data{
    int id;
    LL v[2];
}d[SIZE];
void solve(VPII &AA,LL base){
    LL ret=0;
    REP(i,SZ(AA)){
        d[i].id=AA[i].F;
        d[i].v[0]=d[i].v[1]=0;
        d[i].v[from[AA[i].F]&1]=(LL)AA[i].S*AA[i].F%MOD;
    }
    int dn=SZ(AA);
    while(d[0].id!=1&&dn!=1){
        int nxt_dn=0;
        for(int i=0;i<dn;i++,nxt_dn++){
            if(i+1<dn&&d[i].id/2==d[i+1].id/2){
                REP(j,2)ADD(ret,d[i].v[j]*d[i+1].v[j^1]%MOD*base%MOD*(d[i].id/2));
                d[nxt_dn].id=d[i].id/2;
                REP(j,2)d[nxt_dn].v[j]=(d[i].v[j]+d[i+1].v[j])%MOD;
                i++;
            } else{
                d[nxt_dn].id=d[i].id/2;
                REP(j,2)d[nxt_dn].v[j]=d[i].v[j];
            }
        }
        dn=nxt_dn;
        REP(i,dn)REP(j,2)d[i].v[j]=d[i].v[j]*d[i].id%MOD;
    }
    ADD(an,ret);
}
void dfs(int x){
    if(SZ(top_tree[x])){
        from[top_tree[x][0].F]=x;
        top_tree[x][0].S=x;
        return;
    }
    dfs(2*x);
    dfs(2*x+1);
    top_tree[x].resize(SZ(top_tree[x*2])+SZ(top_tree[x*2+1]));
    merge(ALL(top_tree[2*x]),ALL(top_tree[2*x+1]),top_tree[x].begin());
    solve(top_tree[x],x);
    REP(i,SZ(top_tree[x])){
        from[top_tree[x][i].F]=x;
        top_tree[x][i].S=(LL)top_tree[x][i].S*x%MOD;
    }
}
void solve() {
    int h;
    R(h);
    REPP(i,1<<(h-1),1<<h){
        int x;
        R(x);
        top_tree[i].PB(MP(x+(1<<(h-1))-1,1));
    }
    dfs(1);
    W(an);
}
int main(){
#define MULTITEST 0
#if MULTITEST    
    CASET{
        solve();
    }
#else
    solve();
#endif
    return 0;
}
