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>
#define rep(i,a) for(int i=(int)0;i<(int)a;++i)
#define rrep(i,a) for(int i=(int)a-1;i>=0;--i)
#define REP(i,a,b) for(int i=(int)a;i<(int)b;++i)
#define RREP(i,a,b) for(int i=(int)a-1;i>=b;--i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
typedef std::vector<int> vi;
typedef std::vector<std::vector<int>> vvi;
typedef std::vector<long long> vl;
typedef std::vector<std::vector<long long>> vvl;
using ll=long long;
constexpr ll mod = 1e9 + 7;
constexpr ll INF = 1LL << 60;
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}


ll gcd(ll n, ll m) {
    ll tmp;
    while (m!=0) {
        tmp = n % m;
        n = m;
        m = tmp;
    }
    return n;
}
 
ll lcm(ll n, ll m) {
    return abs(n) / gcd(n, m)*abs(m);//gl=xy
}

using namespace std;

vector<vector<int>>g;
vector<int> ans;
vector<int> dp,a;
const auto inf = numeric_limits<int>::max();

void dfs(int v,int p){
	auto y = lower_bound(all(dp), a[v]);
	int val = *y;
	*y = a[v];
	ans[v] = lower_bound(all(dp),inf)-dp.begin();
	for (auto x : g[v])
	{
		if (ans[x] == -1)
		{
			dfs(x, v);
		}
	}
	*y = val;
}

void solve()
{
  int n;
  cin >> n;
  a.resize(n);
  rep(i, n) cin >> a[i];
  g.resize(n);
  ans.resize(n,-1);
  dp.resize(n, inf);
  rep(i, n - 1)
  {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      g[u].eb(v);
      g[v].eb(u);
  }
  dfs(0, -1);
	rep(i,n){
		cout << ans[i]<< "\n";
	}
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout<<fixed<<setprecision(15);
    solve();
    return 0;
}