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 <cstdio>
#include <queue>

const int N=200005;

int n,p[N],f[N][20],Log[N];

int min(int x,int y){return p[x]<p[y]?x:y;}

int find(int x,int y)
{
    int z=Log[y-x+1];
    return min(f[x][z],f[y-(1<<z)+1][z]);
}

struct Rec
{
    int l,r,x,y;
    Rec(int a=0,int b=0):l(a),r(b),x(find(a,b)),y(find(x+1,b+1)){}
    inline bool operator<(const Rec &o)const{return p[x]>p[o.x];}
};

std::priority_queue<Rec> q;

int main()
{
    scanf("%d\n",&n);
    p[0]=n+1;Log[0]=-1;
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&p[i]);
        f[i][0]=f[i][1]=i;
        Log[i]=Log[i>>1]+1;
    }
    for (int i=n;i;i--)
        for (int j=2;i+(1<<j)-1<=n+1;j++)
            f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);
    q.push(Rec(1,n));
    for (;!q.empty();)
    {
        Rec a=q.top();q.pop();
        printf("%d %d ",p[a.x],p[a.y]);
        if (a.l<a.x) q.push(Rec(a.l,a.x-1));
        if (a.x+1<a.y) q.push(Rec(a.x+1,a.y-1));
        if (a.y<a.r) q.push(Rec(a.y+1,a.r));
    }
}
