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;

const int N = 2e5 + 10;
vector<int> G[N];

vector<int> a;
vector<int> ans;

void dfs(int v, int p, vector<int> &curr, int curr_ans) {
    
    int val = a[v];
    int pos = lower_bound(curr.begin(), curr.end(), val) - curr.begin();
    int pre = curr[pos];
    curr[pos] = val;
    ans[v] = max(curr_ans, pos + 1);
    for(int u : G[v]) if(u != p) {
        dfs(u, v, curr, ans[v]);  
    }
    curr[pos] = pre;
}

int main() {
    ios::sync_with_stdio(false);
    int n; cin >> n;
    a.resize(n); for(int &i : a) cin >> 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);
    }

    ans.resize(n);
    vector<int> start(n, 1e9 + 1000);
    dfs(0, -1, start, 0);
    for(int i : ans) cout << i << "\n";

    return 0;
}