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 INF = 1e9 + 7;
int N;

vector<int> to[200010];
int A[200010];
int ans[200010];
int dp[200010];
bool visited[200010];

void dfs(int now) {
  visited[now] = true;
  auto tmpIdx = lower_bound(dp, dp + N, A[now]);
  int tmpV = *tmpIdx;
  *tmpIdx = A[now];
  ans[now] = lower_bound(dp, dp + N, INF) - dp;

  for (int v : to[now]) {
    if (!visited[v]) {
      dfs(v);
    }
  }

  *tmpIdx = tmpV;
}

int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> A[i];

  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    to[u].push_back(v);
    to[v].push_back(u);
  }

  for (int i = 0; i < 200010; i++) dp[i] = INF;

  dfs(0);
  for (int i = 0; i < N; i++) cout << ans[i] << endl;
}