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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=(1ll<<20);
ll n,a[N],rmq[20][N],pos[N],lg[N];
priority_queue< pair<ll, pair<ll,ll> > >pq;
ll query(ll l,ll r){
    ll mid = (r-l)/2 + 1;
    ll d = lg[mid];
    return min(rmq[d][l] , rmq[d][r-2*((1ll<<d)-1)]);
}
void add(ll l,ll r){
    if(l>r)return;
    ll val = query(l,r-1);
    pq.push({-val,{l,r}});
}
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    cin>>n;
    for(ll i=0;i<n;i++)cin>>a[i],a[i]--,pos[a[i]]=i;
    lg[0]=0;
    for(ll i=1;i<=n;i++){
        ll d = lg[i-1];
        if( (1<<(d+1))<=i )d++;
        lg[i]=d;
    }
    for(ll i=0;i<n;i++)rmq[0][i]=a[i];
    for(ll d=1,l=2;d<20;d++,l*=2){
        for(ll i=0;i<n;i++){
            rmq[d][i] = rmq[d-1][i];
            if(i+l<n)rmq[d][i] = min(rmq[d][i] , rmq[d-1][i+l]);
        }
    }

    add(0,n-1);
    for(ll i=0;i<n/2;i++){
        auto node = pq.top();
        pq.pop();

        ll l = node.second.first;
        ll r = node.second.second;

        ll v1 = query(l,r-1);
        ll p1 = pos[v1];

        ll v2 = query(p1+1,r);
        ll p2 = pos[v2];

        cout<<v1+1<<" "<<v2+1<<" ";
        add(l,p1-1);
        add(p1+1,p2-1);
        add(p2+1,r);
    }
}
