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.

Example

Input

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


Output

1
2
3
3
4
4
5
2
2
3
// Author: HarshKumar

#include <bits/stdc++.h>
using namespace std;
#define ll int64_t
const ll N = 1e6 + 6, M = 1e9 + 7;
vector<ll> g[N], vis(N), a(N), ans(N);
multiset<ll> s;
void dfs(ll v = 0)
{
    vis[v] = 1;
    s.insert(a[v]);
    auto it = s.lower_bound(a[v]);
    it++;
    ll x = 0;
    if (it != s.end())
        s.erase(it),
            x = *it;
    ans[v] = s.size();
    for (ll e : g[v])
        if (!vis[e])
            dfs(e);
    it = s.lower_bound(a[v]);
    if (it != s.end() && *it == a[v])
        s.erase(it);
    if (x)
        s.insert(x);
}
int main()
{
    ios::sync_with_stdio(false), cin.tie(nullptr);
    ll n, u, v;
    cin >> n;
    for (ll i = 0; i < n; i++)
        cin >> a[i];
    for (ll i = 1; i < n; i++)
        cin >> u >> v, u--, v--, g[u].push_back(v), g[v].push_back(u);
    dfs();
    for (ll i = 0; i < n; i++)
        cout << ans[i] << '\n';
}