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;
#define int long long
typedef pair<int,int> pi;
#define f first
#define s second
#define FAST ios_base::sync_with_stdio(0); cin.tie(0);

const int maxn = 200010;

struct node {
	int s,m,e;
	int v;
	node *l, *r;
	node(int _s, int _e) {
		s = _s; e = _e; m = (s+e)/2;
		v = INT_MIN/2;
		if (s != e) {
			l = new node(s,m);
			r = new node(m+1, e);
		}
	}
	
	void up(int x, int newv) {
		if (s == e) {
			v = newv;
			return;
		}
		if (x > m) r -> up(x,newv);
		if (x <= m) l -> up(x,newv);
		v = max(l->v, r->v);
	}
	
	int rmq(int x, int y) {
		if (s == x and e==y) return v;
		if (x > m) return r -> rmq(x,y);
		if (y <= m) return l -> rmq(x,y);
		return max(l->rmq(x,m), r->rmq(m+1,y));
	}
}*root;

vector <int> disc(vector <int> V) {
	vector <int> dis = V;
	sort(dis.begin(),dis.end());
	dis.resize(unique(dis.begin(),dis.end()) - dis.begin());
	for (int i =0;i<(int)V.size();i++) {
		V[i] = lower_bound(dis.begin(),dis.end(),V[i]) - dis.begin() + 1;
	}
	return V;
}

vector <int> adjlist[maxn];
int n;
int ans[maxn];
int A[maxn];

void dfs(int x, int p) {
	
	for (auto i: adjlist[x]) {
		if (i == p) continue;
		int res = root -> rmq(0, A[i] - 1);
		if (res == INT_MIN/2) res = 0;
		root -> up(A[i], res + 1);
		ans[i] = res + 1;
		
		dfs(i,x);
		root -> up(A[i], INT_MIN/2);
	}
	
}

int32_t main() {
	FAST
	cin >> n;
	vector <int> v(n);
	for (int i =0;i<n;i++) {
		cin >> v[i];
	}
	v = disc(v);
	for (int i =1;i<=n;i++) A[i] = v[i-1];
	root = new node(0,n+1);
	
	for (int i =0;i<n-1;i++) {
		int a,b; cin >> a >> b;
		adjlist[a].push_back(b);
		adjlist[b].push_back(a);
	}
	
	root -> up(A[1],1);
	ans[1] = 1;
	
	dfs(1,-1);
	for (int i =1;i<=n;i++) {
		cout << ans[i] << "\n";
	}
}