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;
using ll = long long;
const int INF = 1e9 + 7;





int n;
vector<int> a;
vector<vector<int>> hen;
vector<int> ans;

void dfs(int v, int par, vector<int>& dp) {
    auto itr = lower_bound(dp.begin(), dp.end(), a[v]);
    int past = *itr;
    *itr = a[v];
    ans[v] = lower_bound(dp.begin(), dp.end(), INF) - dp.begin();
    for(auto i : hen[v]) {
        if(i == par)continue;
        dfs(i, v, dp);
    }
    dp[itr - dp.begin()] = past;
}


int main()
{
    cin >> n;
    a.resize(n);
    hen.resize(n);
    ans.resize(n);
    for(int i = 0; i < n; i++)cin >> a[i];
    for(int i = 0; i < n - 1; i++) {
        int s, t; cin >> s >> t; s-- ,t--;
        hen[s].push_back(t); hen[t].push_back(s);
    }
    vector<int> tmp(n + 1, INF);
    dfs(0, -1, tmp);
    for(auto i : ans)cout << i << endl;


}
