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 rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

int n;
int a[200001];
vector<int> g[200001];
int dp[200001],ans[200001];
const int INF=1000100010;
void dfs(int p,int v){
    ans[v]=lower_bound(dp,dp+n,INF)-dp;
    for(auto it:g[v])if(it!=p){
        int pos=lower_bound(dp,dp+n,a[it])-dp;
        int pre=dp[pos];
        dp[pos]=a[it];
        dfs(v,it);
        dp[pos]=pre;
    }
}
int main(){
    cin>>n;
    rep(i,n) cin>>a[i],dp[i]=INF;
    int ui,vi;
    rep(i,n-1){
        cin>>ui>>vi;
        --ui,--vi;
        g[ui].pb(vi);
        g[vi].pb(ui);
    }
    dp[0]=a[0];
    dfs(-1,0);
    FOR(i,0,n) cout<<ans[i]<<endl;
    return 0;
}