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;
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
void in(vector<long long int> &v) {
  for (long long int i = 0; i < v.size(); i++) cin >> v[i];
}
void out(vector<long long int> v) {
  for (long long int i = 0; i < v.size(); i++) cout << v[i] << ' ';
  cout << '\n';
}
void solve() {
  long int n, k;
  cin >> n >> k;
  vector<long long int> a(n), b;
  set<long long int> s;
  in(a);
  for (long long int i = 0; i < n; i++) b.push_back(a[i]);
  sort(b.begin(), b.end());
  vector<long long int> ans;
  for (long long int i = 0; i < n; i++) {
    long long int j = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
    ans.push_back(j);
  }
  vector<long long int> v(n, 0);
  while (k--) {
    long long int x, y;
    cin >> x >> y;
    if (a[x - 1] > a[y - 1]) v[x - 1]--;
    if (a[x - 1] < a[y - 1]) v[y - 1]--;
  }
  for (long long int i = 0; i < ans.size(); i++) {
    ans[i] += v[i];
    if (ans[i] < 0) ans[i] = 0;
  }
  for (auto element : ans) cout << element << " ";
  cout << '\n';
  ;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
