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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i = (m); i <= (n); i++)
#define zep(i,m,n) for(ll i = (m); i < (n); i++)
#define rrep(i,m,n) for(ll i = (m); i >= (n); i--)
#define print(x) cout << (x) << endl;
#define printa(x,m,n) for(int i = (m); i <= n; i++){cout << (x[i]) << " ";} cout<<endl;

ll n, a[200008], x[200008], ans[200008];
vector<vector<ll>> g;

bool f(ll xi, ll aa){
	return x[xi] >= aa;
}

ll cal(ll aa){
	ll ok = n+1;
	ll ng = 0;
	while(ok-ng > 1){
		ll mid = (ok+ng)/2;
		
		if(f(mid, aa)){
			ok = mid;
		}
		else{
			ng = mid;
		}
	}
	return ok; 
}

void dfs(ll at, ll from){
	
	ll ind = cal(a[at]);
	ans[at] = (from == -1)? 1 : max(ans[from], ind);
	
	ll tmp = x[ind];
	x[ind] = a[at];
	
	zep(i, 0, g[at].size()){
		ll nx = g[at][i];
		if(nx != from){
			dfs(nx, at);
		}
	}
	
	x[ind] = tmp;
}

int main(){
	cin.tie(0); ios::sync_with_stdio(false);
	
	cin >> n;
	zep(i, 0, n)cin >> a[i];
	g.resize(n);
	zep(i, 0, n-1){
		ll u, v; cin >> u >> v;
		u--; v--;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	
	rep(i, 1, n)x[i] = INF;
	x[0] = 0;
	
	dfs(0, -1);
	
	zep(i, 0, n){
		print(ans[i])
	}
	return 0;
}