Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

4
3 2 4 1

Sample Output 1

3 1 2 4

The solution above is obtained as follows:

p

q

(3, 2, 4, 1)

()

↓

↓

(3, 1)

(2, 4)

↓

↓

()

(3, 1, 2, 4)

Sample Input 2

2
1 2

Sample Output 2

1 2

Sample Input 3

8
4 6 3 2 8 5 7 1

Sample Output 3

3 1 2 7 4 6 8 5

The solution above is obtained as follows:

p

q

(4, 6, 3, 2, 8, 5, 7, 1)

()

↓

↓

(4, 6, 3, 2, 7, 1)

(8, 5)

↓

↓

(3, 2, 7, 1)

(4, 6, 8, 5)

↓

↓

(3, 1)

(2, 7, 4, 6, 8, 5)

↓

↓

()

(3, 1, 2, 7, 4, 6, 8, 5)

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <iomanip>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vec;
typedef vector<vec> mat;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)
#define pb push_back
#define f first
#define s second
/*
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll max3(ll a, ll b, ll c){return max(a, max(b, c));};
ll min3(ll a, ll b, ll c){return min(a, min(b, c));};
ll max4(ll a, ll b, ll c, ll d){return max(max(a, b), min(c, d));};
ll min4(ll a, ll b, ll c, ll d){return min(min(a, b), min(c, d));};
ll max5(ll a, ll b, ll c, ll d, ll e){return max(max(a, b), max3(c, d, e));};
ll min5(ll a, ll b, ll c, ll d, ll e){return min(min(a, b), min3(c, d, e));};
*/
const ll INFL = 1LL << 60;//10^18 = 2^60
const int INF = 1 << 30;//10^9
ll MOD = 1000000007;
//ll MOD = 998244353;

vector<ll> dy = {0, 0, 1, -1, 1, 1, -1, -1, 0};
vector<ll> dx = {1, -1, 0, 0, 1, -1, 1, -1, 0};

//デバッグ
void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}

ll pow_long(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2) res *= x;
    x *= x;
    k /= 2;
  }
  return res;
}
ll pow_mod(ll x, ll k){
  ll res = 1;
  while(k > 0){
    if(k % 2){
      res *= x; res %= MOD;
    }
    x *= x; x %= MOD;
    k /= 2;
  }
  return res;
}

ll inverse(ll x){return pow_mod(x, MOD - 2);};

ll gcd(ll a, ll b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll x, ll y){return x / gcd(x, y) * y;};

ll kai_mod(ll x){
  if(x == 0) return 1;
  return x * kai_mod(x-1) % MOD;
}

/*
//コンビネーション
const int MAXcomb = 200010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
//facはn!,finvは1/n!
//invは逆元
void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i = 2; i < MAXcomb; i++){
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
ll comb(int n, int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return fac[n] * finv[k] % MOD * finv[n-k] % MOD;
}
*/


const ll N_ = (1 << 18);
ll segOdd[2*N_-1];
//Range Minimum Query
void updateOdd(ll k, ll x){
  k += N_-1;
  segOdd[k] = x;
  while(k > 0){
    k = (k-1)/2;
    segOdd[k] = min(segOdd[2*k+1], segOdd[2*k+2]);
  }
}

ll queryOdd(ll x, ll y, ll k, ll left, ll right){
  if(right <= x || y <= left) return INFL;
  if(x <= left && right <= y) return segOdd[k];
  ll mid = (left + right) / 2;
  ll l = queryOdd(x, y, 2*k+1, left, mid);
  ll r = queryOdd(x, y, 2*k+2, mid, right);
  return min(l, r);
}

ll segEven[2*N_-1];
//Range Minimum Query
void updateEven(ll k, ll x){
  k += N_-1;
  segEven[k] = x;
  while(k > 0){
    k = (k-1)/2;
    segEven[k] = min(segEven[2*k+1], segEven[2*k+2]);
  }
}

ll queryEven(ll x, ll y, ll k, ll left, ll right){
  if(right <= x || y <= left) return INFL;
  if(x <= left && right <= y) return segEven[k];
  ll mid = (left + right) / 2;
  ll l = queryEven(x, y, 2*k+1, left, mid);
  ll r = queryEven(x, y, 2*k+2, mid, right);
  return min(l, r);
}

ll N;
vector<ll> p, q;

vector<pair<ll, ll>> saiki(ll start, ll n){
  if(n == 0) return {};
  if(n == 2){
    return {{p[start], p[start + 1]}};
  }
  ll x, y, x_pos, y_pos;
  if(start % 2){
    x = queryOdd(start, start + n, 0, 0, N_);
    x_pos = q[x];
    y = queryEven(x_pos+1, start + n, 0, 0, N_);
    y_pos = q[y];
  }else{
    x = queryEven(start, start + n, 0, 0, N_);
    x_pos = q[x];
    y = queryOdd(x_pos+1, start + n, 0, 0, N_);
    y_pos = q[y];
  }
  vector<pair<ll, ll>> res = {{x, y}};
  vector<pair<ll, ll>> a1 = saiki(start, x_pos-start);
  vector<pair<ll, ll>> a2 = saiki(x_pos+1, y_pos-x_pos-1);
  vector<pair<ll, ll>> a3 = saiki(y_pos+1, n-(y_pos-start)-1);
  ll at1 = 0, at2 = 0, at3 = 0;
  while(at1 < a1.size() || at2 < a2.size() || at3 < a3.size()){
    pair<ll, ll> A, B, C;
    if(at1 == a1.size()){
      A = {INFL, INFL};
    }else{
      A = a1[at1];
    }
    if(at2 == a2.size()){
      B = {INFL, INFL};
    }else{
      B = a2[at2];
    }
    if(at3 == a3.size()){
      C = {INFL, INFL};
    }else{
      C = a3[at3];
    }
    pair<ll, ll> D;
    if(A <= B && A <= C){
      at1++;
      D = A;
    }else if(B <= A && B <= C){
      at2++;
      D = B;
    }else{
      at3++;
      D = C;
    }
    res.pb(D);
  }
  return res;
}

void solve(){
  rep(i, N){
    updateEven(i, INFL);
    updateOdd(i, INFL);
    if(i % 2) updateOdd(i, p[i]);
    else updateEven(i, p[i]);
  }
  vector<pair<ll, ll>> ans = saiki(0, N);
  rep(i, ans.size()){
    cout << ans[i].f + 1 << " " << ans[i].s + 1;
    if(i == ans.size() - 1) cout << endl;
    else cout << " ";
  }
}

int main(){
  cin >> N;
  p.resize(N);
  q.resize(N);
  rep(i, N){
    cin >> p[i];
    p[i]--;
    q[p[i]] = i;
  }
  solve();
}