// 二分+贪心 O(nlogn)
#include <bits/stdc++.h>
using namespace std;

const int N = 200010;
int n, a[N];
int len, b[N];      // 记录上升子序列
int pos[N], ans[N]; // pos[i]=j 表示a[i]排在b序列中的j位置

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);

    b[0] = -2e9; // 哨兵
    for (int i = 1; i <= n; i++)
    {
        if (b[len] < a[i])
        {
            b[++len] = a[i];
            pos[i] = len;
            // printf("%d %d\n",a[i],len);
        }
        else
        {
            int m = lower_bound(b + 1, b + len + 1, a[i]) - b;
            b[m] = a[i];
            pos[i] = m;
            // printf("%d %d:\n",a[i],m);
        }
    }
    for (int i = n, j = len; i >= 1 && j; i--)
    {
        if (pos[i] == j)
            ans[j--] = a[i];
    }
    for (int i = 1; i <= len; i++)
        printf("%d ", ans[i]);
    return 0;
}