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)

//look at my code
//my code is amazing
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define int long long
#define mod 1000000007 //998244353
#define fast  ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define f(i,n) for(ll i=0;i<n;i++)
#define nl "\n"
#define trace(x) cerr<<#x<<": "<<x<<" "<<endl
#define trace2(x, y) cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z) cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define cout1(a) cout<<a<<nl
#define cout2(a,b) cout<<a<<" "<<b<<nl
#define cout3(a,b,c) cout<<a<<" "<<b<<" "<<c<<nl
#define cout4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<nl
#define vcout(v,i) cout<<v[i].fi<<" "<<v[i].se<<nl
#define x first
#define y second
#define vc vector
#define pb push_back
#define ar array
#define all(a) (a).begin(), (a).end()
#define ms(v,n,x) fill(v,v+n,x);
#define init(c,a) memset(c,a,sizeof(c))
#define pll pair<ll,ll>
#define mll map<ll,ll>
#define sll set<ll>
#define vll vector<ll>
#define vpll vector<pll>
#define maxi(a,b,c) max(a,max(b,c))
#define maxii(a,b,c,d) max(max(a,b),max(c,d))
#define mini(a,b,c) min(a,min(b,c))
#define md(a,b)  ((a%mod)*(b%mod)+mod)%mod 
#define inf 9e18
#define cases ll T;cin>>T;while(T--)
#define BLOCK 500
//const double PI = 3.141592653589793238460;
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
typedef std::complex<double> Complex;
typedef std::valarray<Complex> CArray;
const int N=2e5+5;
ll a[N],ans[N];
vll g[N];
multiset < int > s;
multiset < int > :: iterator it;
void dfs(ll u,ll par=-1){
	s.insert(a[u]);
    
    it = s.lower_bound(a[u]);
    ll del=-1;
	if(++it != s.end()) {
        s.erase(it);
        del=*it;
      }
      
      ans[u]=s.size();
      
      for(auto i:g[u]){
      	if(i==par)
      	continue;
      	dfs(i,u);
      }
      
      if(del!=-1)
      s.insert(del);
      
      s.erase(s.find(a[u]));
}

int32_t main()
{
    
	fast
	ll n;
	cin>>n;
	f(i,n)
	cin>>a[i];
	
	f(i,n-1){
		ll a,b;
		cin>>a>>b;
		a--;b--;
		g[a].pb(b);
		g[b].pb(a);
		
		
	}
	
	dfs(0);
	f(i,n)
	cout<<ans[i]<<nl;

	return 0;
    
}