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 rep(i,n) for (int (i)=(0);(i)<(int)(n);++(i))
using ll = long long;
using namespace std;

#define INF ((1<<30)-1)
#define LLINF (1LL<<60)
#define EPS (1e-10)

vector<int> a, ans;
vector<int> g[201010];

int dp[201010];

void dfs(int v, int p=-1) {
    int i = lower_bound(dp, dp + 201010, a[v]) - dp;
    int old = dp[i];
    dp[i] = a[v];
    ans[v] = i;
    if (p != -1) ans[v] = max(ans[v], ans[p]);
    for (int u : g[v]) {
        if (u == p) continue;
        dfs(u, v);
    }
    dp[i] = old;
}

int main() {
    int n;
    cin >> n;
    a = ans = vector<int>(n);
    rep(i, n) cin >> a[i];
    rep(i, n-1) {
        int x, y;
        cin >> x >> y;
        x--; y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    fill(dp, dp+201010, INF);
    dp[0] = -INF;
    dfs(0);

    rep(i, n) cout << ans[i] << endl;
}
