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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define pb push_back
#define mp make_pair
#define SPEED ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define MOD 1000000007
int visited[200001],val[200001];
vector<int> a[200001],res,resi;
vector<vector<int>>ans;
vector<vector<int>>ansi;
void dfs(int node)
{
    visited[node]=1;
    res.pb(val[node]);
    resi.pb(node);
    ans.pb(res);
    ansi.pb(resi);
    for(int i=0;i<a[node].size();++i)
    {
        if(visited[a[node][i]]==0)
        {
            dfs(a[node][i]);
            res.pop_back();
            resi.pop_back();
        }
    }
}
int CeilIndex(vector<int> v, int l, int r, int key) 
{ 
    while (r - l > 1) { 
        int m = l + (r - l) / 2; 
        if (v[m] >= key) 
            r = m; 
        else
            l = m; 
    } 
  
    return r; 
} 
  
int lis(vector<int> v) 
{ 
    if (v.size() == 0) 
        return 0; 
    vector<int> tail(v.size(), 0); 
    int length = 1; 
    tail[0] = v[0]; 
    for (int i = 1; i < v.size(); i++) 
    { 
        if (v[i] < tail[0]) 
            tail[0] = v[i]; 
        else if (v[i] > tail[length - 1]) 
            tail[length++] = v[i]; 
        else
            tail[CeilIndex(tail, -1, length - 1, v[i])] = v[i]; 
    } 
  
    return length; 
} 
signed main()
{
    SPEED;   
    int n,count=1,i;
    cin>>n;
    for(i=1;i<=n;++i)
        cin>>val[i];
    for(int i=0;i<n-1;++i)
    {
        int u,v;
        cin>>u>>v;
        a[u].pb(v);
        a[v].pb(u);
    }
    dfs(1);
    int fin[n+1];
    for(i=0;i<ansi.size();++i)
    {
        int sz=ansi[i].size();
        fin[ansi[i][sz-1]]=lis(ans[i]);
    }
    for(i=1;i<=n;++i)
        cout<<fin[i]<<endl;
}