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 <iostream>
#include <cstdio>
#include <vector>
#include <stack>
#define PR pair<int,int>
#define F first
#define S second
#define N 200005
using namespace std;

int n, sz, a[N], v[N], d[N];
vector<int> gr[N];
stack<PR> st;

void f(int p) {
	if(v[p]) return;
	int k = lower_bound(d, d+sz, a[p]) - d;
	if(k == sz) sz++;
	v[p] = sz;
	st.push({k, d[k]});
	d[k] = a[p];
	for(auto x:gr[p]) f(x);
	PR t = st.top(); st.pop();
	d[t.F] = t.S;
	if(t.S==1e9) sz--;
}

int main() {
	int i, t1, t2;
	cin >> n;
	for(i=1; i<=n; i++) scanf("%d", &a[i]);
	for(i=0; i<n-1; i++) {
		scanf("%d%d", &t1, &t2);
		gr[t1].push_back(t2);
		gr[t2].push_back(t1);
	}
	for(i=0; i<n; i++) d[i] = 1e9;
	f(1);
	for(i=1; i<=n; i++) printf("%d\n", v[i]);
	return 0;
}