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)

#pragma GCC target ("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<unordered_map>
#include<climits>
#include<fstream>
#include<complex>
#include<time.h>
#include<cassert>
#include<functional>
#include<numeric>
#include<tuple>
using namespace std;
using ll = long long;
using ld = long double;
#define int long long
#define all(a) (a).begin(),(a).end()
#define fs first
#define sc second
#define xx first
#define yy second.first
#define zz second.second
#define H pair<int, int>
#define P pair<int, pair<int, int>>
#define Q(i,j,k) mkp(i,mkp(j,k))
#define rng(i,s,n) for(int i = (s) ; i < (n) ; i++)
#define rep(i,n) rng(i, 0, (n))
#define mkp make_pair
#define vec vector
#define pb emplace_back
#define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end())
#define getidx(b,i) lower_bound(all(b),(i))-b.begin()
#define ssp(i,n) (i==(int)(n)-1?"\n":" ")
#define ctoi(c) (int)(c-'0')
#define itoc(c) (char)(c+'0')
#define pp(x,y) pb(H{x,y})
#define ppp(x,y,z) pb(Q(x,y,z))
#define cyes printf("Yes\n")
#define cno printf("No\n")
#define cdf(n) int quetimes_=(n);rep(qq123_,quetimes_)
#define gcj printf("Case #%lld: ",qq123_+1)
#define readv(a,n) a.resize(n,0);rep(i,(n)) a[i]=read()
//#define endl "\n"
constexpr int mod = 1e9 + 7;
constexpr int Mod = 998244353;
constexpr int MXN = 500000 + 100;
constexpr ld EPS = 1e-10;
constexpr ll inf = 3 * 1e18;
constexpr int Inf = 15 * 1e8;
const vec<int>dx{ -1,1,0,0 }, dy{ 0,0,-1,1 };
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
ll read() { ll u, k = scanf("%lld", &u); return u; }
string reads() { string s; cin >> s; return s; }
H readh(bool g = 0) { H u; int k = scanf("%lld %lld", &u.fs, &u.sc); if (g) u.fs--, u.sc--; return u; }
void printh(H t) { printf("%lld %lld\n", t.fs, t.sc); }
bool inarea(H t, int h, int w) { return 0 <= t.fs && t.fs < h && 0 <= t.sc && t.sc < w; }
ll gcd(ll i, ll j) { return j ? gcd(j, i % j) : i; }
ll lcm(ll i, ll j) { return i / gcd(i, j) * j; }
ll mod_pow(ll x, ll n, ll p = mod) {
	ll res = 1; x %= p;
	while (n > 0) {
		if (n & 1) res = res * x % p;
		x = x * x % p;
		n >>= 1;
	}
	return res;
}//x^n%p
ll bitcount(ll x) {
	int sum = 0; for (int i = 0; i < 60; i++)if ((1ll << i) & x) sum++;
	return sum;
}
constexpr int fn_ = 1000005;
ll fact_[fn_], comp_[fn_];
ll comb(ll x, ll y, ll Mod = mod) {
	if (!fact_[0]) {
		fact_[0] = 1; comp_[0] = 1;
		for (int i = 1; i < fn_; i++)
			fact_[i] = fact_[i - 1] * i % Mod;
		comp_[fn_ - 1] = mod_pow(fact_[fn_ - 1], Mod - 2, Mod);
		for (int i = fn_ - 2; i > 0; i--)
			comp_[i] = comp_[i + 1] * (i + 1) % Mod;
	}
	if (x < y) return 0;
	return fact_[x] * comp_[x - y] % Mod * comp_[y] % Mod;
}
//--------------------------------------------------------------
class RMQ {
#define H pair<int, int>
#define fs first
#define sc second
#define ssiz 1000000
	int size, zero, m;
	H dat[ssiz];
	bool changed[ssiz];
public:
	void init(int siz, int zer) {
		size = siz, zero = zer;
		m = 1;
		while (m < size) m *= 2;
		for (int i = 0; i < ssiz; i++) dat[i] = H{ zero,zero };
	}
	int query(int a, int b) {
		return query_(0, a, b, 0, m);
	}//0-indexed
	void update(int a, int b, int x) {
		update_(0, a, b, 0, m, x);
	}//0-indexed
private:
	void eval(int i, int l, int r) {
		if (!changed[i]) return;
		if (r - l > 1) {
			dat[i * 2 + 1].sc = (dat[i * 2 + 2].sc = dat[i].sc);
			changed[i * 2 + 1] = (changed[i * 2 + 2] = 1);
		}
		dat[i].fs = dat[i].sc;
		changed[i] = 0;
	}
	int query_(int i, int a, int b, int l, int r) {
		eval(i, l, r);
		if (r <= a || b <= l) return zero;
		if (a <= l && r <= b) return dat[i].fs;
		if (zero > 0) return min(query_(i * 2 + 1, a, b, l, (l + r) / 2), query_(i * 2 + 2, a, b, (l + r) / 2, r));
		else return max(query_(i * 2 + 1, a, b, l, (l + r) / 2), query_(i * 2 + 2, a, b, (l + r) / 2, r));
	}
	void update_(int i, int a, int b, int l, int r, int x) {
		eval(i, l, r);
		if (r <= a || b <= l) return;
		if (a <= l && r <= b) {
			dat[i].sc = x;
			changed[i] = 1;
			eval(i, l, r);
			return;
		}
		update_(i * 2 + 1, a, b, l, (l + r) / 2, x);
		update_(i * 2 + 2, a, b, (l + r) / 2, r, x);
		if (zero > 0) dat[i].fs = min(dat[i * 2 + 1].fs, dat[i * 2 + 2].fs);
		else dat[i].fs = max(dat[i * 2 + 1].fs, dat[i * 2 + 2].fs);
	}
};//size, zero
int n;
vec<int>a;
vec<int>e[300000];
RMQ rmq;
int dp[300000];
void solve(int x, int p) {
	dp[x] = rmq.query(0, a[x]) + 1;
	int gg = rmq.query(a[x], a[x] + 1);
	rmq.update(a[x], a[x] + 1, dp[x]);
	for (auto g : e[x]) {
		if (g != p) solve(g, x);
	}
	rmq.update(a[x], a[x] + 1, gg);
}
signed main() {
	cin >> n;
	vec<int>x;
	rep(i, n) {
		a.pb(read());
		x.pb(a[i]);
	}
	crdcomp(x);
	rep(i, n) a[i] = getidx(x, a[i]);
	rep(i, n - 1) {
		int u, v; cin >> u >> v;
		u--; v--;
		e[u].pb(v); e[v].pb(u);
	}
	rmq.init(n, 0);
	solve(0, -1);
	rep(i, n) cout << dp[i] << endl;
}