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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

vector<int> adj[200010];
int n, dp[800010], A[200010];

void build(int a[], int v, int tl, int tr) {
    if (tl == tr) {
        dp[v] = a[tl];
    } else {
        int tm = (tl + tr) / 2;
        build(a, v*2, tl, tm);
        build(a, v*2+1, tm+1, tr);
        dp[v] = max(dp[v*2], dp[v*2+1]);
    }
}

int sum(int v, int tl, int tr, int l, int r) {
    if (l > r)
        return 0;
    if (l == tl && r == tr) {
        return dp[v];
    }
    int tm = (tl + tr) / 2;
    return max(sum(v*2, tl, tm, l, min(r, tm))
           , sum(v*2+1, tm+1, tr, max(l, tm+1), r));
}

void update(int v, int tl, int tr, int pos, int new_val) {
    if (tl == tr) {
        dp[v] = new_val;
    } else {
        int tm = (tl + tr) / 2;
        if (pos <= tm)
            update(v*2, tl, tm, pos, new_val);
        else
            update(v*2+1, tm+1, tr, pos, new_val);
        dp[v] = max(dp[v*2], dp[v*2+1]);
    }
}

int lis[200010];

void dfs(int u, int p, int l) {
    //update dp pos
    int mx = sum(1, 1, n, 1, A[u] - 1);
    update(1, 1, n, l, mx + 1);

//    cout << "u " << u << endl;
//    cout << "l " << l << endl;
//    cout << "mx " << mx << endl;
//    cout << endl;

    lis[u] = mx + 1;

    for(auto v : adj[u]) {
        if(v == p) continue;
        dfs(v, u, l + 1);
    }
    //rollback dp pos
    update(1, 1, n, l, 0);
}

int main() {
    ios_base::sync_with_stdio(false);

    cin >> n;

    for(int i = 1; i <= n; i++) {
        cin >> A[i];
    }

    for(int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;

        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs(1, 0, 1);
    build(dp, 1, 1, n);

    for(int i = 1; i <= n; i++) cout << lis[i] << endl;
    cout << endl;
}