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>
#define ll long long
using namespace std;
vector <ll> tree[200008];
ll arr[200009];
int ans[200009];
vector <ll> dp(200009,LLONG_MAX);
void dfs(int cur,int p)
{
    int pos=lower_bound(dp.begin(),dp.end(),arr[cur])-dp.begin();
    ll x=dp[pos];
    dp[pos]=arr[cur];

    for(ll i:tree[cur])
    {
        if(i!=p)
            dfs(i,cur);
    }
    ans[cur]=lower_bound(dp.begin(),dp.end(),LLONG_MAX)-dp.begin();
    dp[pos]=x;


}
int main()
{
    ll i,j,k,l,m,n;
    cin>>n;
    for(i=0;i<n;i++)
        cin>>arr[i];
    for(i=0;i<n-1;i++)
    {
        ll a,b;
        cin>>a>>b;
        a--;
        b--;
        tree[a].push_back(b);
        tree[b].push_back(a);
    }
    dfs(0,-1);
    for(i=0;i<n;i++)
        cout<<ans[i]<<endl;


}
