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 ll long long
#define INF 1000000000
#define pb push_back
vector<ll> ans(2000001,0);
vector<int> dp(2000001,INF);
vector<int> val(2000001);
vector<ll> adj[200001];

void dfs(ll u,ll p)
{
    //cout<<"U::"<<u<<"P::"<<p<<endl;
    ll k=lower_bound(dp.begin(),dp.end(),val[u])-dp.begin();
    ll temp=dp[k];
    dp[k]=val[u];
    ans[u]=max(k,ans[p]);
    for(int j=0;j<adj[u].size();j++)
    {
        if(adj[u][j]==p)continue;
        dfs(adj[u][j],u);
    }
    dp[k]=temp;
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)cin>>val[i];
    for(int i=1;i<n;i++)
    {
        ll u,v;
        cin>>u>>v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    dp[0]=-INF;
    dfs(1,0);
    for(int i=1;i<=n;i++)
        cout<<ans[i]<<endl;
}
