Codeforces Round #353 (Div. 2) D. Tree Construction

During the programming classes Vasya was assigned a difficult problem. However, he doesn't know how to code and was unable to find the solution in the Internet, so he asks you to help.

You are given a sequence a, consisting ofn distinct integers, that is used to construct the binary search tree. Below is the formal description of the construction process.

1. First element a₁ becomes the root of the tree.
2. Elements a₂, a₃, ..., a_n are added one by one. To add elementa_i one needs to traverse the tree starting from the root and using the following rules:
   1. The pointer to the current node is set to the root.
   2. If a_i is greater than the value in the current node, then its right child becomes the current node. Otherwise, the left child of the current node becomes the new current node.
   3. If at some point there is no required child, the new node is created, it is assigned valuea_i and becomes the corresponding child of the current node.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the length of the sequencea.

The second line contains n distinct integersa_i (1 ≤ a_i ≤ 10⁹) — the sequencea itself.

Output

Output n - 1 integers. For all i > 1 print the value written in the node that is the parent of the node with valuea_i in it.

Examples

Input

31 2 3

Output

1 2

Input

54 2 3 1 6

Output

4 2 2 4

Note

Picture below represents the tree obtained in the first sample.

Picture below represents the tree obtained in the second sample.

 

题意：给你一个数列，让你以此构造一个排序二叉树，输出每个点的父亲。

分析：再次被智商压制，排序二叉树中一个点向左找到的第一个点（左下或左上）一定是之前序列中出现的最后一个小于等于当前节点权值得数，右边同理。

#include <set> #include <cstdio>#include <iostream>#include <unordered_map>using namespace std;set <int> f;unordered_map <int,int> l,r;int n,a[100005];int main(){cin.sync_with_stdio(false);cin>>n;for(int i = 1;i <= n;i++){cin>>a[i];if(i == 1) f.insert(a[i]);else {auto it = f.lower_bound(a[i]);if(it == f.end()) r[*--it] = a[i];else {if(!l[*it]) l[*it] = a[i];else r[*--it] = a[i];}cout<<*it<<" ";f.insert(a[i]);}}}