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<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<utility>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<functional>
#include<math.h>
using namespace std;
#define N (1000000000+7)
#define M (998244353)
//#define INF 1e16
typedef long long ll;
typedef pair<int,vector<ll> > P;
typedef pair<P,ll>Q;
 
const int MAX_N = 200005;
const int INF = 1001001001;
vector<int>g[MAX_N];
vector<int>a,ans;
int dp[MAX_N];

void dfs(int v,int p=-1){
    int index = lower_bound(dp,dp+MAX_N,a[v])-dp;
    ll old = dp[index];
    dp[index]=a[v];
    ans[v] = index;
    if(p!=-1)ans[v]=max(ans[v],ans[p]);
    for(int j=0;j<(ll)g[v].size();j++){
        int next = g[v][j];
        if(next == p)continue;
        dfs(next,v);
    }
    dp[index] = old;
    return;
}

int main(void){
    ll n;
  cin>>n;
   a = ans = vector<int>(n);
  for(ll i=0;i<n;i++)cin>>a[i];
  for(ll i=0;i<n-1;i++){
    ll u,v;
    cin>>u>>v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for(ll i=0;i<MAX_N;i++)dp[i]=INF;
  dp[0]=-INF;
  dfs(0);
  for(int i=0;i<n;i++)cout<<ans[i]<<endl;
  return 0;
}