#include <iostream>
using namespace std;
int n, arr[200005], used[400005];
//0: operationnya pasti ga kepake, but the number still ther
//1: number is not there, operationnya dah kepake
//2: nothing, untouched
const int inf = 1e9 + 69;
int query(int id){
int rs = inf;
for(; id; id >>= 1){
if(~used[id] & 1) rs = min(rs, arr[id]);
if(!used[id]) break;
if((id & 1) && used[id ^ 1]){
rs = min(rs, arr[id ^ 1]);
if(used[id ^ 1] == 1) break;
}
}
return rs;
}
void erase(int id, int val){
for(; id; id >>= 1){
if(arr[id] == val){
used[id] = 0;
break;
}
if(id & 1){
if(arr[id ^ 1] == val){
used[id ^ 1] = used[id] = 1;
break;
}else used[id ^ 1] = 0;
}
used[id] = 1;
}
}
#define parl (i << 1)
#define parr (parl | 1)
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n;
for(int i = 1; i <= n; i ++){
cin >> arr[i];
used[i] = 2;
}
used[1] = 0;
for(int i = 1; i <= n; i ++){
int rs = query(i);
int upl = (parl <= n ? arr[parl] : inf);
int upr = (parr <= n ? arr[parr] : inf);
int v = min(rs, min(upl, upr));
cout << v;
if(i == n) cout << endl;
else cout << " ";
if(rs == v){
used[parl] = used[parr] = 0;
erase(i, rs);
}else if(upl == v){
used[parl] = 1;
used[parr] = 0;
}else if(upr == v){
used[parr] = 1;
}
}
return 0;
}
