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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
typedef int64_t Int;
typedef pair<Int,Int> p;

int N; 
vector<Int> A(200010);
vector<vector<int>> G(200010);

vector<vector<int>> path(200010);
vector<int> depth(200010,-1);
vector<p> ans(200010,p(0,0));

void solve(int to, int v){
    if(ans[v].second < A[to]) ans[to] = p(ans[v].first+1, A[to]);
    else {
        if(A[v] > A[to]) ans[to] = p(ans[v].first, A[to]);
        else ans[to] = p(ans[v].first, A[v]);
    }
}

void setup(int v){
    for(int to : G[v]){
        if(depth[to] == -1){
            //cout << v << " " << to << endl;
            depth[to] = depth[v]+1;
            path[to] = path[v];
            path[to].push_back(A[to]);
            solve(to, v);
            setup(to);
        }
    }
}


int main(){
    cin >> N;
    rep(i,N) cin >> A[i];

    rep(i,N-1){
        int u, v; cin >> u >> v; u--; v--;
        G[u].push_back(v);
        G[v].push_back(u);
    }

    depth[0] = 0;
    path[0].push_back(A[0]);
    ans[0] = p(A[0],1);
    setup(0);

    rep(i,N) cout << ans[i].first << endl;
}