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>
std::map<int, int> rankn, rank;
std::vector<int> arg[200010];
int r[200010];
std::pair<int, int> sumrank[200010];
int main() {
  int n, k;
  std::cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    std::cin >> r[i];
    rankn[r[i]]++;
  }
  for (int i = 0; i < k; i++) {
    int x, y;
    std::cin >> x >> y;
    arg[x].push_back(y);
    arg[y].push_back(x);
  }
  int rankcnt = 1;
  for (auto i = rankn.begin(); i != rankn.end(); i++) {
    int first = i->first;
    int second = i->second;
    sumrank[rankcnt] =
        std::make_pair(first, sumrank[rankcnt - 1].second + second);
    rank[first] = rankcnt;
    rankcnt++;
  }
  for (int i = 1; i <= n; i++) {
    int ans = sumrank[rank[r[i]] - 1].second;
    for (int j : arg[i]) {
      if (r[j] < r[i]) {
        ans--;
      }
    }
    std::cout << ans << " ";
  }
  std::cout << std::endl;
  return 0;
}
