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;
#define ll long long
#define ld long double
#define INF 0x3f3f3f3f
const int N=2e5 + 5;
const ll mod = 1e9 + 7;

int n, u, v;
ll val[N], LIS[N], len, ans[N];
vector<int> grid[N];

void dfs(int c, int p) {
    int idx = lower_bound(LIS + 1, LIS + n + 1, val[c]) - LIS;
    int temp = LIS[idx];
    LIS[idx] = val[c];
    ans[c] = max(ans[p], (ll)idx);
    for(auto next : grid[c]) {
        if(next != p) {
            dfs(next, c);
        }
    }
    LIS[idx] = temp;
}

int main()
{
    ios_base::sync_with_stdio(0),cin.tie(NULL),cout.tie(NULL);
    cin >> n;
    for(int i=1;i<=n;i++) {
        cin >> val[i];
        LIS[i] = INF;
    }
    for(int i=1;i<n;i++) {
        cin >> u >> v;
        grid[u].push_back(v);
        grid[v].push_back(u);
    }
    dfs(1, 0);
    for(int i=1;i<=n;i++) cout << ans[i] << endl;
    return 0;
}
