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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <vector>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <numeric>
#include <string>
#include <sstream>
#include <cmath>
#include <cstring>
#include <assert.h>
#include <utility>
#include <tuple>
#include <array>
#include <bitset>
#include <cstdlib>

using int64 = long long;

using namespace std;

int N;
vector<int> as, answers, table;
vector<vector<int>> tree;

void DFS(int u, int p) {
    // tableを二分探索
    int i = lower_bound(begin(table), end(table), as[u]) - begin(table);
    int old_value = table[i];

    table[i] = as[u];
    answers[u] = lower_bound(begin(table), end(table), (1 << 30)) - begin(table);

    for (int c : tree[u]) {
        if (c == p) continue;
        DFS(c, u);
    }

    table[i] = old_value;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    cin >> N;
    as.resize(N);
    for (int i = 0; i < N; i++) {
        cin >> as[i];
    }

    tree.resize(N);
    for (int i = 0; i < N - 1; i++) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        tree[u].push_back(v);
        tree[v].push_back(u);
    }

    answers.resize(N);
    table.resize(N, (1 << 30));
    DFS(0, -1);

    for (int i = 0; i < N; i++) {
        cout << answers[i] << endl;
    }

    return 0;
}