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;

#define rep(i,a,n) for(int i=a;i<n;++i)
#define fi first
#define se second
#define pb push_back
#define mp make_pair

const int N=2e5+5;
const int inf=2e9;
int a[2][N];
int pos[N];
int f[2][N][32];
struct node{
    int val,l,r;
    bool friend operator<(node a,node b){
        return a.val>b.val;
    }
};

void rmq_init(int v,int n){
    for(int i=0;i<n;++i) {
        f[v][i][0]=a[v][i];
    }
    for(int j=1;(1<<j)<=n;++j){
        for(int i=0;i+(1<<j)-1<n;++i){
            f[v][i][j]=min(f[v][i][j-1],f[v][i+(1<<(j-1))][j-1]);
        }
    }
}
int query(int v,int l,int r){
    int k=0;
    while(1<<(k+1)<=(r-l+1)) ++k;
    return min(f[v][l][k],f[v][r-(1<<k)+1][k]);
}
int main(){
    ios::sync_with_stdio(0);
    int n,t;
    cin>>n;
    rep(i,0,n){
        cin>>t;
        pos[t]=i;
        a[i&1][i]=t;
        a[(i&1)^1][i]=inf;
    }
    rmq_init(0,n);
    rmq_init(1,n);

    priority_queue<node>q;
    t=query(0,0,n-1);
    q.push({t,0,n-1});
    while(!q.empty()){
        node t=q.top();
        int v1=t.val;
        q.pop();
        int v2=query((pos[v1]&1)^1,pos[v1]+1,t.r);
        cout<<v1<<' '<<v2<<' ';
        v1=pos[v1];
        v2=pos[v2];
        if(v1!=t.l){
            q.push({query(t.l&1,t.l,v1-1),t.l,v1-1});
        }
        if(v1+1!=v2){
            q.push({query((v1+1)&1,v1+1,v2-1),v1+1,v2-1});
        }
        if(v2!=t.r){
            q.push({query((v2+1)&1,v2+1,t.r),v2+1,t.r});
        }
    }
    return 0;

}
