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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#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 = 1e9;
vector<int> a, dp, ans;
vector<int> g[200005];

void dfs(int u, int p = -1){
  int i = lower_bound(dp.begin(),dp.end(),a[u]) - dp.begin();
  int temp = dp[i];
  dp[i] = a[u];
  ans[u] = p==-1 ? i : max(i,ans[p]);
  for(int v : g[u]) if(v!=p) dfs(v,u);
  dp[i] = temp;
}

int main() {
  int n;
  cin >> n;
  a = vector<int>(n);
  rep(i,n) cin >> a[i];
  rep(i,n-1){
    int u,v;
    cin >> u >> v;
    --u; --v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }

  dp = vector<int>(n+2, INF);
  dp[0] = -INF;
  ans = vector<int>(n);

  dfs(0);
  rep(i,n) cout << ans[i] << endl;
}
