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;

using ll = long long;
#define rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = (n); i > 0; i--)
#define all(n) n.begin(), n.end()
#define print(n) cout << n << endl

ll n;
vector<ll> a(200000),dp(200000,1<<30),ans(200000);
vector<set<ll>> e(200000);

int dfs(ll v,ll prev){
    ll pos = lower_bound(all(dp),a[v])-dp.begin();
    ll temp = dp[pos];
    dp[pos] = a[v];
    ans[v]=(lower_bound(all(dp),1<<30)-dp.begin());
    
    for(ll i:e[v]) if(i!=prev) dfs(i,v);
    dp[pos] = temp;
    return 0;
}

int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    
    ll u,v;

    cin >> n;
    rep(i,n) cin >> a[i];
    rep(i,n-1){
        cin >> u >> v;
        e[u-1].insert(v-1);
        e[v-1].insert(u-1);
    }
    dfs(0,0);

    rep(i,n) print(ans[i]);
    return 0;
}