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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define fi first
#define se second
#define pb push_back
#define forn(i,n) for(int i=0;i<(n);i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define forr(i,n) for(int i=n;i>=0;i--)
#define all(x) x.begin(), x.end()
const LL MAXN = 2e5 + 5;

void fio(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}

vector <LL> ar(MAXN);
vector<vector<LL> > adj(MAXN);

LL max(LL a, LL b){
	return a>b?a:b;
}

LL t[2*MAXN + 10];

void modify(LL p, LL value) {  
  for (t[p += MAXN] = value; p > 1; p >>= 1) t[p>>1] = max(t[p], t[p^1]);
}

LL query(int l, int r) { 
  LL res = 0;
  for (l += MAXN, r += MAXN; l < r; l >>= 1, r >>= 1) {
    if (l&1) res = max(res,t[l++]);
    if (r&1) res = max(res,t[--r]);
  }
  return res;
}

LL an[MAXN];

void dfs(int u,int p){
	LL old = query(ar[u], ar[u]+1);
	LL	x = query(0, ar[u]); 
	LL value = x + 1; 
	an[u] = value;
	modify(ar[u], value);
	for(auto i:adj[u]){
		if(i!=p){
			dfs(i,u);
		}
	}
	modify(ar[u], old);
}

int main(){
  fio();
  int n; cin>>n;
  vector<LL> v(n);
  set<LL> s;
  forn(i,n) {
  	cin>>v[i];
  	s.insert(v[i]);
  }
  LL in = 0;
  map<LL,LL> mp;
  for(auto i:s){
  	mp[i] = ++in;
  }
  forn(i,n){
  	ar[i+1] = mp[v[i]];
  }

  forn(i,n-1){
  	int u, v; cin>>u>>v;
  	adj[v].pb(u);
  	adj[u].pb(v);
  }
  dfs(1,0);
  for1(i,n) cout<<an[i]<<endl;

  return 0;
}