#include <bits/stdc++.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0;}
template<class T>bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0;}

using ll = long long;
using R = long double;
const R EPS = 1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r) {return (r > EPS) - (r < -EPS);}
inline R sq(R x) {return sqrt(max(x, 0.0L));}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// Problem Specific Parameter:

// Description: セグメント木　点更新 区間クエリ
// TimeComplexity: 初期化$\mathcal{O}(n \log n)$ 更新とクエリ$\mathcal{O}(\log n)$
// Verifyed: AOJ DSL_2_A

struct Segment_tree {
	using T = ll;
	const T out = (1LL << 31) - 1;

	int n;
	vector<T> data;

	inline T vmerge(T l, T r) {return min(l, r);}

	Segment_tree(int n): n(n) {data.assign(2 * n, out);}

	void update(int p, T x) { // set value at position p
		for (data[p += n] = x; p > 1; p >>= 1) {
			data[p >> 1] = vmerge(data[p], data[p ^ 1]);
		}
	}

	T query(int l, int r) {  // sum on interval [l, r)
		T resl = out, resr = out;
		for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
			if (l & 1) resl = vmerge(data[l++], resl);
			if (r & 1) resr = vmerge(resr, data[--r]);
		}
		return vmerge(resl, resr);
	}
};

const int limit = 200010;
Segment_tree odd(limit);
Segment_tree even(limit);
int idx[limit];

int main(void) {
	int n;
	cin >> n;

	vector<int> p(n);
	rep(i, n) {
		cin >> p[i];
		if (i & 1)
			odd.update(i, p[i]);
		else
			even.update(i, p[i]);
		idx[p[i]] = i;
	}

	vector<int> ans;
	using state = tuple<int, int, int, int>;
	priority_queue<state, vector<state>, greater<state>> q;


	{
		int x = even.query(0, n);
		int y = odd.query(idx[x], n);
		q.push(state(x, y, 0, n));
	}

	while (!q.empty()) {
		int x, y, l, r;
		tie(x, y, l, r) = q.top(); q.pop();
		//cerr << x << " " << y << " " << l << " " << r << " " << ans.size() << endl;
		ans.push_back(x);
		ans.push_back(y);

		int xi = idx[x], yi = idx[y];

		if (xi - l >= 2) {
			int nx, ny;
			if (l & 1) {
				nx = odd.query(l, xi);
				ny = even.query(idx[nx], xi);
			} else {
				nx = even.query(l, xi);
				ny = odd.query(idx[nx], xi);
			}
			q.push(state(nx, ny, l, xi));
		}

		if (yi - xi - 1 >= 2) {
			int nx, ny;
			if ((xi + 1) & 1) {
				nx = odd.query(xi + 1, yi);
				ny = even.query(idx[nx], yi);
			} else {
				nx = even.query(xi + 1, yi);
				ny = odd.query(idx[nx], yi);
			}
			q.push(state(nx, ny, xi + 1, yi));
		}

		if (r - yi - 1 >= 2) {
			int nx, ny;
			if ((yi + 1) & 1) {
				nx = odd.query(yi + 1 , r);
				ny = even.query(idx[nx], r);
			} else {
				nx = even.query(yi + 1, r);
				ny = odd.query(idx[nx], r);
			}
			q.push(state(nx, ny, yi + 1, r));
		}

		//cerr << endl;
	}

	rep(i, n) cout << (i ? " " : "") << ans[i];
	cout << endl;

	return 0;
}

Measured memory_bytes: 10868