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 <algorithm>
#include <queue>
using namespace std;
const int maxn = 2e5+7;
int n,a[maxn],log[maxn],f[2][18][maxn],pos[maxn];

int rmq(int k,int l,int r){
    
    int j = log[r-l+1];
    return min(f[k][j][l],f[k][j][r-(1<<j)+1]);
}
pair<int,int> cal(int l,int r){
    int x = rmq(l&1,l,r);
    int y = rmq((pos[x]+1)&1,pos[x]+1,r);
    
    return {-x,-y};
}
int main(){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        pos[a[i]]=i;
    }
    for(int i=2;i<=n;i++){
        log[i]=log[i>>1]+1;
    }
    for(int l=0;l<2;l++){
        for(int i=0;i<n;i++){
            f[l][0][i]=(i%2==l)?a[i]:1<<30;
        }
        for(int k=1;1<<k<=n;k++){
            for(int j=0;j+(1<<k)-1<n;j++){
                f[l][k][j]=min(f[l][k-1][j],f[l][k-1][j+(1 << k - 1)]);
            }
        }
    }
    priority_queue< pair<pair<int,int> ,pair<int,int> > > Q;
    Q.push({cal(0,n-1),{0,n-1}});
    while(!Q.empty()){
        auto it = Q.top();Q.pop();
        int x = -it.first.first;
        int y = -it.first.second;
        printf("%d %d ",x,y);
        int l = it.second.first;
        int r = it.second.second;
        x = pos[x],y = pos[y];
        if(l<x-1){
            Q.push({cal(l,x-1),{l,x-1}});
        }
        if(x+1<y-1){
            Q.push({cal(x+1,y-1),{x+1,y-1}});
        }
        if(y+1<r){
            Q.push({cal(y+1,r),{y+1,r}});
        }
    }
    
}