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;

typedef long long ll;

const int N = 200005;
int n;
int A[N];
vector<int> edges[N];
vector<int> V(N, INT_MAX);
int res[N];

void dfs(int par, int u) {
    for (auto v : edges[u]) {
        if (v == par) continue;
        int pos = lower_bound(V.begin(), V.end(), A[v]) - V.begin();

        res[v] = max(res[u], pos + 1);

        int temp = V[pos];
        V[pos] = A[v];
        dfs(u, v);
        V[pos] = temp;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> A[i];
    }
    int u, v;
    for (int i = 0; i < n - 1; ++i) {
        cin >> u >> v;
        edges[u].push_back(v);
        edges[v].push_back(u);
    }
    res[1] = 1;
    V[0] = A[1];
    dfs(0, 1);

    for (int i = 1; i <= n; ++i) {
        cout << res[i] << endl;
    }

    return 0;
}

