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 ull unsigned long long
#define endl '\n'
#define ll long long
const int N = 2e5 + 5;
ll a[N], n;
vector<ll> node[N];
multiset<ll> s[N];
ll dp[N];
void dfs(ll u, ll p){
        ll l = 1, h = n;
        while (l <= h){
                ll mid = (l + h) >> 1;
                if (*s[mid].begin() >= a[u]) h = mid - 1;
                else l = mid + 1;
        }
        dp[u] = h + 1;
        s[h + 1].insert(a[u]);
        for (auto j : node[u]){
                if (j != p) {
                        dfs(j, u);
                }
        }
        auto it = s[h + 1].lower_bound(a[u]);
        s[h + 1].erase(it);
}
int main(){
        ios_base::sync_with_stdio(false);
        cin.tie(0);
        cout.tie(0);
        if (fopen("WALKING.INP", "r")){
                freopen("WALKING.INP", "r", stdin);
                freopen("WALKING.OUT", "w", stdout);
        }
        cin >> n;
        for (int i = 1; i <= n; i++){
                cin >> a[i];
        }
        s[0].insert(0);
        for (int i = 1; i <= n; i++){
                s[i].insert(1e18);
        }
        for (int i = 1; i < n; i++){
                ll u, v;
                cin >> u >> v;
                node[u].push_back(v);
                node[v].push_back(u);
        }
        dfs(1, 0);
        for (int i = 1; i <= n; i++) cout << dp[i] << endl;
}