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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

const int INF = 1001001001;
int A[200005];
vector<int> E[200005];
int d[200005];
bool visited[200005];
int lis[200005];
int n;

void dfs(int u) {
    visited[u] = true;
    int idx = lower_bound(lis, lis+n+1, A[u]) - lis;
    int old_num = lis[idx];
    lis[idx] = A[u];
    d[u] = (lower_bound(lis, lis+n+1, INF) - lis) - 1;
    for (int v: E[u]) {
        if (visited[v]) continue;
        dfs(v);
    }
    lis[idx] = old_num;
    return;
}

int main() {
    cin >> n;
    rep(i, n) cin >> A[i];
    rep(i, n-1) {
        int u, v;
        cin >> u >> v;
        --u; --v;
        E[u].push_back(v);
        E[v].push_back(u);
    }
    rep(i, 200005) lis[i] = INF;
    lis[0] = 0;
    dfs(0);
    rep(i, n) {
        cout << d[i] << endl;
    }
    return 0;
}