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;
typedef long long ll;
typedef pair<ll,ll> P;
const ll mod=1e9+7,INF=mod*mod*3;//M_PI
#define rep(i,N) for(ll i=0; i<(N); i++)
#define rep1(i,N) for(ll i=1; i<(N); i++)
#define pb push_back

ll N;
ll a[200005];
ll dp[200005];
ll ans[200005];
vector<ll> vec[200005];
void dfs(ll in,ll nm,ll noINF){
 	ll x=upper_bound(dp,dp+N+1,a[nm])-dp;
 	ll be=dp[x];
    if(dp[x-1]<a[nm]) dp[x]=a[nm];
    if(dp[x]!=INF) noINF=max(noINF,x);
    if(dp[in]!=INF) noINF=in;
    ans[nm]=noINF; 
    for(auto &i:vec[nm]){
    	if(ans[i]==INF) dfs(in+1,i,noINF);
    }
    dp[x]=be;
	return;
}

int main(){

cin.tie(0);
ios::sync_with_stdio(false);

cin>>N;
rep1(i,N+1) cin>>a[i];
rep(i,N-1){
    ll u,v; 
    cin>>u>>v;
    vec[u].pb(v);
    vec[v].pb(u);
}
fill(ans,ans+N+1,INF);
fill(dp,dp+N+1,INF);
dp[0]=-1;
dfs(1,1,0);
rep1(i,N+1) cout<<ans[i]<<endl;

}
    
