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
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 87;
int n, a[N], ans[N];
vector<int> t = {0};
vector<int> g[N];
void dfs(int u, int p)
{
    int pb = 0, pj = -1, x;
    if (a[u] > t.back()) {
        t.push_back(a[u]);
        pb = 1;
    } else {
        pj = lower_bound(t.begin(), t.end(), a[u]) - t.begin();
        x = t[pj];
        t[pj] = a[u];
    }
    ans[u] = t.size() - 1;
    for (int v : g[u])
        if (v != p)
            dfs(v, u);
    if (pb)
        t.pop_back();
    else
        t[pj] = x;
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; ++i)
        cin >> a[i];
    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, 1);
    for (int i = 1; i <= n; ++i)
        cout << ans[i] << '\n';
}
