Let N be a positive even number.

We have a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N). Snuke is constructing another permutation of (1, 2, ..., N), q, following the procedure below.

First, let q be an empty sequence. Then, perform the following operation until p becomes empty:

* Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.



When p becomes empty, q will be a permutation of (1, 2, ..., N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

* N is an even number.
* 2 ≤ N ≤ 2 × 10^5
* p is a permutation of (1, 2, ..., N).

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the lexicographically smallest permutation, with spaces in between.

Examples

Input

4
3 2 4 1


Output

3 1 2 4


Input

2
1 2


Output

1 2


Input

8
4 6 3 2 8 5 7 1


Output

3 1 2 7 4 6 8 5
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <map>
#include <vector>
#include <math.h>
#include <algorithm>
#include <queue>
#include <set>
#include <tuple>
using namespace std;

#define rep(i,a) for(int i=0; i<a; i++)
#define rrep(i,a) for(int i=a; i>=0; i--)
#define rep1(i,a) for(int i=1; i<=a; i++)
#define cout1(a) cout << a << endl;
#define cout2(a,b) cout << a << " " << b << endl;
#define cout3(a,b,c) cout << a << " " << b << " " << c << endl;
#define cout4(a,b,c,d) cout << a << " " << b << " " << c << " " << d << endl;
#define mem(a,n) memset( a, n, sizeof(a))
#define all(a) a.begin(),a.end()

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> V;
typedef vector<V> VV;
typedef vector<VV> VVV;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LLINF = 1e18;
static const double pi = 3.141592653589793;

struct st{int l, r, x, y;};
int n, a[200009], m[200009][20];
priority_queue<st> pq;
bool operator<(st p, st q){return a[p.x]>a[q.x];}

int mn(int p, int q){return a[p]<a[q]?p:q;}

int f(int l,int r){
    int j=0;
    while(1<<(j+1)<=r-l+1) j++;
    return mn(m[l][j],m[r-(1<<j)+1][j]);
}

void add(int l,int r){
    int x, y;
    if(l>r) return;
    x=f(l,r);
    y=x+1==r?r:mn(f(x+1,r-1),r);
    pq.push({l,r,x,y});
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    cin>>n;
    rep(i,n) cin>>a[i], m[i][1]=i;
    for(int j=2;;j++){
        int k=1<<j;
        if(k>n) break;
        for(int i=0;i+k<=n;i++) m[i][j]=mn(m[i][j-1],m[i+k/2][j-1]);
    }
    for(add(0,n-1);!pq.empty();){
        st t=pq.top(); pq.pop();
        cout<<a[t.x]<<" "<<a[t.y]<<" ";
        add(t.l,t.x-1);
        add(t.x+1,t.y-1);
        add(t.y+1,t.r);
    }
    cout<<endl;
}