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;
const int inf = 0x3f3f3f3f;
int n;
int v[200005];
int dp[200005];
int f[200005];
vector<vector<int>> g;
void dfs(int p, int u) {
    int pos = lower_bound(f, f+n, v[p]) - f;
    int t = f[pos];
    f[pos] = v[p]; 
    dp[p] = lower_bound(f, f+n, inf) - f;
    for(auto i:g[p]) {
        if(i!=u)
        dfs(i, p);
    }
    f[pos] = t;
}
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    cin>>n;
    g.resize(n+1);
    for(int i=0;i<n;i++) cin>>v[i];
    fill(f,f+n+2,inf);
    for(int i=1;i<n;i++) {
        int x, y;
        cin>>x>>y;
        x--; y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs(0, -1);
    for(int i=0;i<n;i++) cout<<dp[i]<<endl;
    return 0;
}