//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define enl printf("\n")
#define case(t) printf("Case #%d: ", (t))
#define ni(n) scanf("%d", &(n))
#define nl(n) scanf("%lld", &(n))
#define nai(a, n) for (int i = 0; i < (n); i++) ni(a[i])
#define nal(a, n) for (int i = 0; i < (n); i++) nl(a[i])
#define pri(n) printf("%d\n", (n))
#define prl(n) printf("%lld\n", (n))
#define pii pair<int, int>
#define pil pair<int, long long>
#define pll pair<long long, long long>
#define vii vector<pii>
#define vil vector<pil>
#define vll vector<pll>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef cc_hash_table<int,int,hash<int>> ht;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
const double pi = acos(-1);
const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const int MAXN = 2e5 + 5;
const double eps = 1e-9;
int a[MAXN*4];
int n;
map<pii,int> dp;
int ok(int k, int val) {
if (dp.find({k, val}) != dp.end()) return dp[{k, val}];
if (k*2 > n) return k;
if (k*2+1 > n) {
if (val > a[k*2]) return k*2;
return k;
}
if (val < min(a[k*2], a[k*2+1]))
return dp[{k, val}] = k;
else if (a[k*2] < min(val, a[k*2+1]))
return dp[{k, val}] = ok(k*2, val);
int mi = min(val, a[k*2]);
if (ok(k*2, mi) < ok(k*2+1, mi)) {
if (mi == val)
return dp[{k, val}] = ok(k*2, val);
return dp[{k, val}] = ok(k*2+1, val);
} else {
if (mi == a[k*2])
return dp[{k, val}] = ok(k*2, val);
return dp[{k, val}] = ok(k*2+1, val);
}
}
void solve(int k) {
if (k * 2 > n) return;
if (k * 2 + 1 > n) {
if (a[k] > a[2 * k]) swap(a[k], a[2 * k]);
return;
}
int mi = min(a[k], min(a[k*2],a[k*2+1]));
if (mi == a[k*2]) {
swap(a[k], a[k*2]);
} else if (mi == a[k*2+1]) {
int mi = min(a[k], a[k*2]), mx = max(a[k], a[k*2]);
a[k] = a[k*2+1];
if (ok(2*k,mi) < ok(2*k+1,mi))
a[k*2] = mi, a[k*2+1] = mx;
else
a[k*2+1] = mi, a[k*2] = mx;
}
solve(k + 1);
}
int main() {
ni(n);
for (int i = 1; i <= n; i++)
ni(a[i]);
solve(1);
for (int i = 1; i <= n; i++)
printf("%d ", a[i]);
enl;
}
