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>//Carefully Crafted by hetp111
using namespace std;
#define int long long
#define double long double
#define all(v) (v).begin(),(v).end()
#define vi vector<int>
#define vvi vector<vi>
#define pii pair<int,int>
#define vii vector<pii>
#define MOD (1000000007)
#define MOD2 (998244353)
#define MOD3 (1000000009)
#define PI acos(-1)
#define eps (1e-8)
#define INF (1e18)
#define FASTER ios_base::sync_with_stdio(0);cin.tie(0)
template<class A,class B>ostream&operator<<(ostream&out,const pair<A,B>&a){return out<<"("<<a.first<<","<<a.second<<")";}
template<class A>ostream&operator<<(ostream&out,const vector<A>&a){for(const A &it:a)out<<it<<" ";return out;}
template<class A,class B>istream&operator>>(istream&in,pair<A,B>&a){return in>>a.first>>a.second;}
template<class A>istream&operator>>(istream&in,vector<A>&a){for(A &i:a)in>>i;return in;}
ifstream cinn("input.txt");ofstream coutt("output.txt");

int n,a[200005],ans[200005];
vi adj[200005];
multiset<int> st;

void dfs(int cur,int par){
    
    
    auto it=st.lower_bound(a[cur]);
    int del=0;
    if(it!=st.end()){
        del=*it;
        st.erase(st.find(del));
        st.insert(a[cur]);
    }else{
        st.insert(a[cur]);
    }
    
    ans[cur]=st.size();
    
    for(int ch:adj[cur]){
        if(ch==par)continue;
        dfs(ch,cur);
    }
    
    if(del)st.insert(del);
    st.erase(st.find(a[cur]));

}

signed main(){
    FASTER;
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<n;i++){
        int a,b;cin>>a>>b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++)cout<<ans[i]<<"\n";
}