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 <cstdio>
#include <vector>
typedef std::vector<int> vin;
const int INFN = 1e9+7;
const int MAXN = 2e5+5;
int A[MAXN];
vin G[MAXN];
int L[MAXN];      // LIS stack
int up[MAXN];     // DFS parent
int N;

int ret[MAXN];

void dfs(int u) {
  int pos = std::lower_bound(L, L + N, A[u]) - L;
  int save = L[pos];
  L[pos] = A[u];
  
  ret[u]  = std::lower_bound(L, L + N, INFN) - L;
  for (int v: G[u]) {
    if (v != up[u]) {
      up[v] = u;
      dfs(v);
    }
  }
  L[pos] = save;
}

int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &A[i]);
  }
  for (int i = 1; i < N; ++i) {
    int u, v;
    scanf(" %d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 0; i < N; ++i) {
    L[i] = INFN;
  }
  up[1] = 1;
  dfs(1);
  for (int i = 1; i <= N; ++i) {
    printf("%d\n", ret[i]);
  }
  return 0;
}