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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <fstream>
#include <vector>
#include  <map>
#include <algorithm>
#include <iostream>

// BE CAREFUL WITH HASH TABLE & UNORDERED MAP
// #include <ext/pb_ds/assoc_container.hpp>
// using namespace __gnu_pbds;
using namespace std;

#define MOD 998244353
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define F0R(i,b) FOR(i,0,b)
#define RFO(i,a,b) for (int i = (b-1); i >= (a); i--)
#define RF0(i,b) RFO(i,0,b)
#define ll long long
#define pii pair<int,int>
#define add(a,b) ((int) (((ll) (a) + (b))%MOD))
#define mul(a,b) ((int) (((ll) (a) * (b))%MOD))
#define PB emplace_back
#define MP make_pair
#define F first
#define S second

// ifstream cin ("cinput.in");
// ofstream cout ("coutput.out");

int ct = 0;
int N, a, b;
int vals [200000];
int ans [200000];
vector<int> graph [200000];
map<int,int> dp;
pii insd [200000];
pii deld [200000];

// ins. if same val exists, should already be optimal

int ins(int node) { // insert and delete suboptimal
    auto it = dp.lower_bound(vals[node]);
    if (it->F != vals[node]) {
        if (it == dp.begin()) {
            insd[node] = MP(vals[node], 1);
            dp[vals[node]] = 1;
        } else if (prev(it)->S < vals[node]) {
            int asdf = prev(it)->S + 1;
            insd[node] = MP(vals[node], asdf);
            dp[vals[node]] = asdf;
        }
        // cout << it->S << '\n';
        if (it != dp.end() && it->S <= prev(it)->S) {
            int sd = it->F;
            int df = it->S;
            deld[node] = MP(sd, df);
            dp.erase(it);
        }
    }
    int sdf = prev(dp.end())->S;
    return sdf;
}

void dfs(int node, int par) {
    ans[node] = ins(node);
    ct++;
    for (int ne: graph[node]) {
        if (ne == par) continue;
        dfs(ne, node);
    }
    if (deld[node].F != -1) dp[deld[node].F] = deld[node].S;
    if (insd[node].F != -1) dp.erase(dp.find(insd[node].F));
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> N;
    F0R(i,N) {
        insd[i] = MP(-1,-1);
        deld[i] = MP(-1,-1);
    }
    F0R(i,N) cin >> vals[i];
    F0R(i,N-1) {
        cin >> a >> b;
        a--; b--;
        graph[a].PB(b);
        graph[b].PB(a);
    }
    dfs(0,-1);
    F0R(i,N) cout << ans[i] << '\n';
}