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;

int n, u, v, a[200000], d[200000];
vector<int> adj[200000];

vector<int> lis = {int(-1e9)};

void dfs(int u) {
    bool inserted = (a[u] > lis.back());
    int idx, replaced;
    if (inserted) {
        lis.push_back(a[u]);
    }
    else {
        auto it = lower_bound(lis.begin(), lis.end(), a[u]);
        idx = it - lis.begin();
        replaced = lis[idx];
        lis[idx] = a[u];
    }

    d[u] = lis.size() - 1;

    for (int v : adj[u]) {
        if (!d[v]) dfs(v);
    }
    if (inserted) lis.pop_back();
    else lis[idx] = replaced;
}

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