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::vector<std::pair<long long, long long>> v;
std::vector<long long> t;
std::map<long long, std::pair<long long, long long>> map;
std::map<long long, long long> map2;
std::vector<std::vector<long long>> adjlist;
struct cmp {
  bool operator()(std::pair<long long, long long> a,
                  std::pair<long long, long long> b) {
    return a.second < b.second;
  }
};
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  long long n, m;
  std::cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    long long a;
    std::cin >> a;
    v.push_back({i + 1, a});
    map2[i + 1] = a;
  }
  std::sort(v.begin(), v.end(), cmp());
  adjlist.assign(n + 1, std::vector<long long>());
  for (long long i = 0; i < m; i++) {
    long long a, b;
    std::cin >> a >> b;
    if (map2[a] > map2[b]) {
      adjlist[a].push_back(b);
    } else if (map2[a] < map2[b]) {
      adjlist[b].push_back(a);
    }
  }
  t.assign(n, 0);
  for (long long i = 0; i < n; i++) {
    std::pair<long long, long long> current = v[i];
    long long id = current.first;
    long long skill = current.second;
    long long should = -1;
    if (map[skill].first == 0) {
      map[skill].first++;
      map[skill].second = i;
      should = i;
    } else {
      should = map[skill].second;
    }
    long long a = should;
    a -= (long long)adjlist[id].size();
    long long ans = (a < 0) ? 0 : a;
    t[id - 1] = ans;
  }
  for (long long i = 0; i < n; i++) {
    std::cout << t[i] << " ";
  }
  return 0;
}
