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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>
using namespace __gnu_pbds;
using namespace std;
typedef tree<
    pair<int, int>,
    null_type,
    less_equal<pair<int, int>>,
    rb_tree_tag,
    tree_order_statistics_node_update>
    ordered_set;
#define int long long
#define double long double
#define all(a) (a).begin(), (a).end()
#define x first
#define y second
#define lb(v, z) lower_bound(all(v), z)
#define ub(v, z) upper_bound(all(v), z)
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define rdp(i, a, b) for (int i = (a); i >= (b); i--)
#define rlt(i, a, b) for (int i = (a); i < (b); i++)
#define cns(z) cout << (z) << ' '
#define cnl(z) cout << (z) << '\n'
#define M1 1000000007
#define M2 998244353
#define MAXN 300005
#define INF (1ll << 60)
#define endl "\n"
#define garr(ip)       \
    for (auto &x : ip) \
        cin >> x;
#define parr(ip)          \
    for (auto &x : ip)    \
        cout << x << " "; \
    cout << endl;
#define nl cout << endl
typedef vector<int> vi;
typedef pair<int, int> pii;
typedef vector<pair<int, int>> vp;

vi a, ans;
vector<vi> g;
deque<int> dq;

void dfs(int s, int p = 0)
{
    auto it = ub(dq, a[s]);
    bool flag = false;
    int num = 0;
    if (it == dq.end())
    {
        dq.push_back(a[s]);
    }
    else
    {
        flag = true;
        num = *it;
        *it = a[s];
    }
    ans[s] = dq.size();
    for (auto it : g[s])
        if (it != p)
            dfs(it, s);
    if (flag)
        *it = num;
    else
        dq.pop_back();
}

void solve(int testCaseNumber = 1)
{
    int n;
    cin >> n;
    a = vi(n + 1);
    ans = vi(n + 1);
    g = vector<vi>(n + 1);
    rep(i, 1, n) cin >> a[i];
    rlt(i, 1, n)
    {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1);
    rep(i, 1, n) cnl(ans[i]);
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    solve();
}