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: Accepted)

#include <bits/stdc++.h>
#define ll long long
#define ls id << 1
#define rs id << 1 | 1
#define mem(array, value, size, type) memset(array, value, ((size) + 5) * sizeof(type))
#define memarray(array, value) memset(array, value, sizeof(array))
#define pb(x) push_back(x)
#define st(x) (1LL << (x))
#define pii pair<int, int>
#define mp(a, b) make_pair((a), (b))
#define Flush fflush(stdout)
using namespace std;
const int N = 500050;
const int inf = 0x3f3f3f3f;
const ll mod = 998244353LL;
clock_t TIME_START, TIME_END;
void program_end()
{
#ifdef ONLINE
    printf("\nTime used: %.6lf(s)\n", ((double)TIME_END - TIME_START) / CLOCKS_PER_SEC);
    system("pause");
#endif
}
int n;
int a[N];
vector<int> e[N];
multiset<int> s;
int tot;
int ans[N];

void dfs(int u, int fa)
{
    auto it = s.lower_bound(a[u]);
    multiset<int> t;
    if (it == s.end())
        tot++;
    else
    {
        t.insert(*it);
        s.erase(it);
    }
    s.insert(a[u]);
    ans[u] = tot;
    for (auto v : e[u])
    {
        if (v == fa)
            continue;
        dfs(v, u);
    }
    s.erase(s.find(a[u]));
    for (auto &i : t)
        s.insert(i);
    tot = ans[fa];
}

void solve()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    for (int i = 1; i < n; ++i)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; ++i)
        printf("%d\n", ans[i]);
}

int main()
{
    TIME_START = clock();
    int Test = 1;
    // cin >> Test;
    while (Test--)
        solve();
    TIME_END = clock();
    program_end();
    return 0;
}