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 a[200005], dp[200005], ans[200005], sz = 1;
vector<int> g[200005];
void dfs(int u, int p){
    int idx = lower_bound(dp, dp+sz, a[u]) - dp;
    int pos = -1, val = 0;
    if(idx == sz) dp[idx] = a[u], sz++;
    else{
        pos = idx, val = dp[idx];
        dp[idx] = a[u];
    }
    ans[u] = sz-1;

    for(int i=0;i<g[u].size();i++){
        int v = g[u][i];
        if(v == p) continue;
        dfs(v, u);
    }

    if(pos != -1) dp[pos] = val;
    else sz--;
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &a[i]);
    for(int i=1;i<=n-1;i++){
        int a, b;
        scanf("%d%d", &a, &b);
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    dp[0] = 0;
    dfs(1, 0);
    for(int i=1;i<=n;i++) printf("%d\n", ans[i]);
}