In BerSoft n programmers work, the programmer i is characterized by a skill r_i.

A programmer a can be a mentor of a programmer b if and only if the skill of the programmer a is strictly greater than the skill of the programmer b (r_a > r_b) and programmers a and b are not in a quarrel.

You are given the skills of each programmers and a list of k pairs of the programmers, which are in a quarrel (pairs are unordered). For each programmer i, find the number of programmers, for which the programmer i can be a mentor.

Input

The first line contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ k ≤ min(2 ⋅ 10^5, (n ⋅ (n - 1))/(2))) — total number of programmers and number of pairs of programmers which are in a quarrel.

The second line contains a sequence of integers r_1, r_2, ..., r_n (1 ≤ r_i ≤ 10^{9}), where r_i equals to the skill of the i-th programmer.

Each of the following k lines contains two distinct integers x, y (1 ≤ x, y ≤ n, x ≠ y) — pair of programmers in a quarrel. The pairs are unordered, it means that if x is in a quarrel with y then y is in a quarrel with x. Guaranteed, that for each pair (x, y) there are no other pairs (x, y) and (y, x) in the input.

Output

Print n integers, the i-th number should be equal to the number of programmers, for which the i-th programmer can be a mentor. Programmers are numbered in the same order that their skills are given in the input.

Examples

Input

4 2
10 4 10 15
1 2
4 3


Output

0 0 1 2 


Input

10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5


Output

5 4 0 5 3 3 9 0 2 5 

Note

In the first example, the first programmer can not be mentor of any other (because only the second programmer has a skill, lower than first programmer skill, but they are in a quarrel). The second programmer can not be mentor of any other programmer, because his skill is minimal among others. The third programmer can be a mentor of the second programmer. The fourth programmer can be a mentor of the first and of the second programmers. He can not be a mentor of the third programmer, because they are in a quarrel.
#include <bits/stdc++.h>
using namespace std;
int N;
pair<int, int> arr[200005];
int ans[200005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int M;
  cin >> N >> M;
  for (int i = 1; i <= N; i++) cin >> arr[i].first, arr[i].second = i;
  sort(arr + 1, arr + 1 + N);
  for (int i = 1; i <= N; i++) {
    int idx =
        lower_bound(arr + 1, arr + 1 + N, pair<int, int>{arr[i].first, 0}) -
        arr;
    ans[arr[i].second] = idx - 1;
  }
  sort(arr + 1, arr + 1 + N, [=](pair<int, int> a, pair<int, int> b) -> bool {
    return a.second < b.second;
  });
  for (int i = 0, u, v; i < M; i++) {
    cin >> u >> v;
    if (arr[u].first > arr[v].first) ans[u]--;
    if (arr[v].first > arr[u].first) ans[v]--;
  }
  for (int i = 1; i <= N; i++) {
    if (i != 1) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}
