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 <algorithm>
#include <vector>
using namespace std;
vector<vector<int> > adj;
vector<int> A, ans, s;

void dfs(int i, int p) {
    int j = lower_bound(s.begin(), s.end(), A[i]) - s.begin();
    int save = s[j];
    s[j] = A[i];
    ans[i] = max(ans[i], j);
    for (int j : adj[i]) {
        if (j != p) { ans[j] = ans[i]; dfs(j, i); }
    }
    s[j] = save;
}

int main() {
    int n; scanf("%d", &n);
    adj.resize(n);
    A.resize(n);
    for (int &a : A) scanf(" %d", &a);
    for (int i = 1; i < n; i++) {
        int u, v; scanf(" %d %d", &u, &v);
        u--; v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    ans.resize(n);
    s.resize(n, 1<<30);
    s[0] = 0;
    dfs(0, -1);
    for (int a : ans) printf("%d\n", a);
}
