Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices, whose i-th edge connects Vertex u_i and Vertex v_i.
Vertex i has an integer a_i written on it.
For every integer k from 1 through N, solve the following problem:

We will make a sequence by lining up the integers written on the vertices along the shortest path from Vertex 1 to Vertex k, in the order they appear. Find the length of the longest increasing subsequence of this sequence.

Here, the longest increasing subsequence of a sequence A of length L is the subsequence A_{i_1} , A_{i_2} , ... , A_{i_M} with the greatest possible value of M such that 1 \leq i_1 < i_2 < ... < i_M \leq L and A_{i_1} < A_{i_2} < ... < A_{i_M}.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq a_i \leq 10^9

1 \leq u_i , v_i \leq N

u_i \neq v_i

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N
u_1 v_1
u_2 v_2
:
u_{N-1} v_{N-1}

Output

Print N lines. The k-th line, print the length of the longest increasing subsequence of the sequence obtained from the shortest path from Vertex 1 to Vertex k.

Sample Input 1

10
1 2 5 3 4 6 7 3 2 4
1 2
2 3
3 4
4 5
3 6
6 7
1 8
8 9
9 10

Sample Output 1

1
2
3
3
4
4
5
2
2
3

For example, the sequence A obtained from the shortest path from Vertex 1 to Vertex 5 is 1,2,5,3,4. Its longest increasing subsequence is A_1, A_2, A_4, A_5, with the length of 4.

Solution in C++ (Clang 10.0.0) (judge verdict: Accepted)

//g++ -std=c++14 test.cpp -o test.out

#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iostream>
#include <iomanip>
#include <vector>
#include <queue>
#include <stack>
#include <map> 
#include <unordered_map> 
#include <utility>
#include <limits.h>
#include <bitset>
#include <set>
using namespace std;
 
#define LL long long int
const LL INF = (1LL<<60);
const int INF_INT = 2147483647-1e6-1;
const LL mod = 1000000007ll;
const int mod_int = 1000000007;

LL N;
LL A[200001];
vector<vector<LL>> adj;

vector<LL> ans;

vector<LL> LIS_array;
bool used[200001] = {};
void dfs(LL now){
	used[now] = 1;
	bool pop_flag = false;
	int insert_index = int(lower_bound(LIS_array.begin(),LIS_array.end(),A[now])-LIS_array.begin());
	LL prev_num;

	if(insert_index==LIS_array.size()){
		pop_flag = true;
		LIS_array.push_back(A[now]);
	}else{
		prev_num = LIS_array[insert_index];
		LIS_array[insert_index] = A[now];
	}

	ans[now] = LIS_array.size();

	// cout << now << endl;
	// for(int i=0;i<LIS_array.size();i++) cout << LIS_array[i] << " ";
	// cout << endl;

	for(LL next:adj[now]){
		if(used[next]) continue;
		//cout << now << "->" << next << endl;
		dfs(next);
		//cout << now << "<-" << next << endl;
	}

	if(pop_flag) LIS_array.pop_back();
	else LIS_array[insert_index] = prev_num;
}

void solve(){
	ans = vector<LL>(N+1);
	dfs(1);
}

int main(){
    cin >> N;
    for(int i=1;i<=N;i++) cin >> A[i];
    adj = vector<vector<LL>>(N+1);
	for(int i=0;i<=N-2;i++){
		LL u,v;cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}

    solve();

    for(int i=1;i<=N;i++) cout << ans[i] << endl;
    return 0;
}