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
#define eps 1e-8
#define INF 0x3f3f3f3f
#define pb push_back
#define endl '\n'
using namespace std;
const ll maxn = 1e6 + 5;
ll n,w[maxn],a,b,LIS[maxn],len;
ll ans[maxn];
vector<int> G[maxn];
void dfs(int x,int f){
    int idx = lower_bound(LIS + 1,LIS + n + 1,w[x]) - LIS;
    int tmp = LIS[idx];
    LIS[idx] = w[x];
    ans[x] = max(ans[f],(ll)idx);
    for(int i = 0;i < G[x].size(); i++){
        int endd = G[x][i];
        if(endd != f)
            dfs(endd,x);
    }
    LIS[idx] = tmp;
}
int main()
{
    cin >> n;
    int a,b;
    for(int i = 1;i <= n; i++) cin >> w[i],LIS[i] = INF;;
    for(int i = 1; i < n; i++)
        cin >> a >> b,G[a].pb(b),G[b].pb(a);
    dfs(1,0);
    for(int i = 1; i <= n; i++)
        cout << ans[i] << endl;
    return 0;
}

