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 INF=1e9+9,MAX_N=2e5+10;

int N;
int a[MAX_N];
vector<int> G[MAX_N];
int ans[MAX_N];
vector<int> dp;

void dfs(int v,int p){
    int nxt=lower_bound(dp.begin(),dp.end(),a[v])-dp.begin();
    int tmp=dp[nxt];
    dp[nxt]=a[v];
    ans[v]=lower_bound(dp.begin(),dp.end(),INF)-dp.begin();
    for (int u:G[v]){
        if (u==p) continue;
        dfs(u,v);
    }
    dp[nxt]=tmp;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    for (int i=0;i<N;++i) cin >> a[i];
    for (int i=0;i<N-1;++i){
        int u,v; cin >> u >> v; --u,--v;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }
    for (int i=0;i<N;++i) dp.emplace_back(INF);
    dfs(0,-1);
    for (int i=0;i<N;++i) cout << ans[i] << '\n';
}