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>
typedef long long ll;
using namespace std;
using Graph = vector<vector<int>>;

Graph g;
vector<int> a, ans;
const int N = 2000005;
const int INF = 1000000000;
int dp[N];

void dfs(int v, int p=-1) {
  int i = lower_bound(dp, dp + N, a[v]) - dp;
  int old = dp[i];
  dp[i] = a[v];
  ans[v] = i;
  if (p != -1) ans[v] = max(ans[v], ans[p]);
  for(int u : g[v]) {
    if (u == p) continue;
    dfs(u, v);
  }
  dp[i] = old;
}

int main() {
  int n;
  cin >> n;
  g.resize(n);
  a.resize(n);
  ans.resize(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--;
    g[u].push_back(v);
    g[v].push_back(u);
  }

  dp[0] = -INF;
  for(int i = 1; i < N; ++i) {
    dp[i] = INF;
  }
  dfs(0);
  for(int i = 0; i < n; ++i) {
    cout << ans[i] << endl;
  }
  return 0;
}