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 for_(i,a,b) for(int i=a;i<b;++i)
#define ALL(a) a.begin(),a.end()
#define INIT(a,b) memset(a,b,sizeof(a))
using lint = long long int;

int N, a[200005];
int dp[200005], ans[200005];
vector<int> adj[200005];

void rec(int v, int p) {
    int i = lower_bound(dp, dp+N, a[v]) - dp;
    int prev = dp[i];

    // 更新
    dp[i] = a[v];
    ans[v] = max(ans[p], i+1);

    // 子へ
    for (int u : adj[v]) if (u != p) rec(u, v);

    // 戻す
    dp[i] = prev;
}

int main() {
    cin >> N;
    for_(i,0,N) cin >> a[i];
    for_(i,0,N-1) {
        int u, v;
        cin >> u >> v;
        --u; --v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }    

    for_(i,0,N+1) dp[i] = 1L << 30;
    ans[N] = 0;
    rec(0, N);
    for_(i,0,N) cout << ans[i] << endl;
}