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<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;

vector<int>g[200010];
bool visited[200010];
int ans[200010];

int main(void){
  int n;
  cin>>n;
  vector<ll>a(n);
  for(int i=0;i<n;i++)cin>>a[i];
  for(int i=0;i<n-1;i++){
    int u,v;
    cin>>u>>v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  vector<ll>dp(n,INF);
  auto itr = lower_bound(dp.begin(),dp.end(),a[0]);
  *itr = a[0];
  ans[0] = lower_bound(dp.begin(),dp.end(),INF)-dp.begin();
  queue<P>que;
  que.push(P(0,dp));
  while(!que.empty()){
    P p = que.front();
    que.pop();
    ll now = p.first;
    visited[now]=true;
    for(ll i=0;i<(ll)g[now].size();i++){
      int next =g[now][i];
      if(!visited[next]){
        vector<ll>dp1;
        dp1 = p.second;
        auto itr = lower_bound(dp1.begin(),dp1.end(),a[next]);
        *itr = a[next];
        ans[next]=lower_bound(dp1.begin(),dp1.end(),INF)-dp1.begin();
        que.push(P(next,dp1));
      }
    }
  }
  for(int i=0;i<n;i++)cout<<ans[i]<<endl;
  return 0;
}