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: Runtime Error)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <utility>
#include <complex>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <deque>
#include <tuple>
#include <bitset>
#include <limits>
#include <algorithm>
#include <array>
#include <random>
#include <complex>
#include <regex>
using namespace std;
typedef long double ld;
typedef long long ll;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef complex<ld> compd;
#define quickIO()	{cin.tie(0);	cout.sync_with_stdio(false);}
#define reach(i,a)	for(auto i:a)
#define rep(i,n)	for(int i=0;i<((int)n);i++)
#define REP(i,n)	for(int i=0;i<=((int)n);i++)
#define srep(i,a,n)	for(int i=a;i<((int)n);i++)
#define SREP(i,a,n)	for(int i=a;i<=((int)n);i++)
#define rrep(i,n)	for(int i=n-1;i>=0;i--)
#define RREP(i,n)	for(int i=n;i>=0;i--)
#define all(a)	(a).begin(),(a).end()
#define mp(a,b)	make_pair(a,b)
#define mt	make_tuple
#define pb	push_back
template<typename T> istream& operator >> (istream& is, vector<T>& vec) {
	for (T& x : vec)	is >> x;
	return is;
}
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
	os << "[";
	rep(i, vec.size())	os << (i ? ", " : "") << vec[i];
	os << "]";
	return os;
}
template<typename T> ostream& operator << (ostream& os, set<T>& s) {
	os << "{";
	for (T x : s)	os << x << ", ";
	os << "}";
	return os;
}
template<typename T> istream& operator >> (istream& is, pair<T, T>& p) {
	is >> p.first >> p.second;
	return is;
}
template<typename T> ostream& operator << (ostream& os, pair<T, T>& p) {
	os << p.first << " " << p.second;
	return os;
}
template<typename T> bool operator < (vector<T>& a, vector<T>& b) {
	rep(i, a.size()) {
		if (i == b.size())	return false;
		if (a[i] < b[i])	return true;
		else if (a[i] > b[i])	return false;
	}
	return false;
}
template<typename T> bool operator > (vector<T>& a, vector<T>& b) {
	rep(i, a.size()) {
		if (i == b.size())	return true;
		if (a[i] < b[i])	return false;
		else if (a[i] > b[i])	return true;
	}
	return false;
}
int bitcnt(ll x) {
	x = ((x & 0xAAAAAAAAAAAAAAAA) >> 1) + (x & 0x5555555555555555);
	x = ((x & 0xCCCCCCCCCCCCCCCC) >> 2) + (x & 0x3333333333333333);
	x = ((x & 0xF0F0F0F0F0F0F0F0) >> 4) + (x & 0x0F0F0F0F0F0F0F0F);
	x = ((x & 0xFF00FF00FF00FF00) >> 8) + (x & 0x00FF00FF00FF00FF);
	x = ((x & 0xFFFF0000FFFF0000) >> 16) + (x & 0x0000FFFF0000FFFF);
	x = ((x & 0xFFFFFFFF00000000) >> 32) + (x & 0x00000000FFFFFFFF);
	return x;
}
int bitcnt(int x) {
	x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555);
	x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333);
	x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F);
	x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF);
	x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF);
	return x;
}
ll sqrtll(ll x) {
	ll left = 0, right = x;
	rep(i, 100) {
		ll mid = (left + right) >> 1;
		if (mid * mid <= x)	left = mid;
		else	right = mid;
	}
	return left;
}
ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a % b);
}
#define debug(x)	printf("Case #%d: ", x)
#undef DEBUG
const ld infl = 1e100;
const ll mod = 1e9 + 7;
const ld eps = 1e-9;
const ll inf = 1e15;
const int dx[] = { 1,0,-1,0,1, -1 };
const int dy[] = { 0,1,0,-1,1, -1 };

vint dp[200010];
vint edge[200010];
vint a;

void dfs(int pos, int parent) {
	// insert
	dp[pos] = dp[parent];
	assert(dp[pos].size() < 15);
	auto target_pos = lower_bound(all(dp[pos]), a[pos]);
	if (target_pos == dp[pos].end())	dp[pos].push_back(a[pos]);
	else	*target_pos = a[pos];
	// to next node
	for (int to : edge[pos]) {
		if (parent == to)	continue;
		dfs(to, pos);
	}
}

int main() {
	int n;	cin >> n;
	a.resize(n);	cin >> a;
	srep(i, 1, n) {
		int u, v;	cin >> u >> v;	u--;	v--;
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	dp[0].push_back(a[0]);
	for (int to : edge[0]) {
		dfs(to, 0);
	}
	rep(i, n) {
		cout << dp[i].size() << endl;
	}
	return 0;
}