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;

const int N = 2e5+2;

//compress A[i] first
//so that we can use LIS via Segment tree

long long s;
long long e;
long long val;
long long segtree[N << 2];

void update(int P , int L , int R){
     if(L > e || R < s){
        return;
     } else if(s <= L && R <= e) {
        segtree[P] = val;
        return;
     }
     int mid = (L + R) >> 1;
     update( (P << 1) | 0 , L , mid + 0);
     update( (P << 1) | 1 , mid + 1 , R);
     //segtree[P] = segtree[P << 1] + segtree[P << 1 | 1];
     segtree[P] = max(segtree[P << 1] , segtree[P << 1 | 1]);
}

long long query(int P , int L , int R){
    if(L > e || R < s){
        return 0;
    } else if(s <= L && R <= e){
        return segtree[P];
    } else {
        int mid = (L + R) >> 1;
        long long a = query( (P << 1) | 0 , L , mid + 0);
        long long b = query( (P << 1) | 1 , mid + 1 , R);
        return max(a , b);
    }
}

int n;
int a[N];
int ans[N];
map < int , int > to;
vector < int > adj[N];

void dfs(int u , int p){
    //cout << u << "\n";
    s = 1 , e = a[u] - 1;

    ans[u] = query(1 , 1 , n) + 1;
    e++;
    s = e;
    val = ans[u];
    update(1 , 1 ,  n);

    ans[u] = max(ans[u] , ans[p]);

    for(auto v : adj[u]){
        if(v == p)
            continue;
        dfs(v , u);
    }

    s = a[u];
    e = a[u];
    val = 0;
    update(1 , 1 , n);
}

int main(){
    scanf("%d" , &n);
    vector < int > all;
    for(int i = 1; i <= n; i++){
        scanf("%d" , &a[i]);
        all.push_back(a[i]);
    }
    int idx = 2;
    sort(all.begin() , all.end());
    for(int i = 0; i < n; i++){
        if(!i){
            to[all[i]] = idx;
            continue;
        }
        if(all[i] == all[i - 1]){
            to[all[i]] = to[all[i - 1]];
        } else {
            to[all[i]] = ++idx;
        }
    }
    for(int i = 1; i <= n; i++){
        a[i] = to[a[i]];
    }

    for(int i = 1; i < n; i++){
        int u , v;
        scanf("%d%d" , &u , &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs(1 , 0);
    for(int i = 1; i <= n; i++){
        printf("%d\n" , ans[i]);
    }
}