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;
vector<pair<int, int> > a, b;
map<int, int> cnt;
int n, k, ans[200005];
inline int bs(int key) {
  int kq = b.size();
  for (int l = 0, r = b.size() - 1; l <= r;) {
    int m = (l + r) >> 1;
    if (b[m].first < key) {
      kq = m;
      l = m + 1;
    } else
      r = m - 1;
  }
  return kq;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  a.push_back(make_pair(0, 0));
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    if (!cnt[x]) b.push_back(make_pair(x, 0));
    cnt[x]++;
    a.push_back(make_pair(x, i));
  }
  for (int i = 1; i <= k; ++i) {
    int u, v;
    cin >> u >> v;
    if (a[u].first > a[v].first) ans[a[u].second]--;
    if (a[u].first < a[v].first) ans[a[v].second]--;
  }
  sort(b.begin(), b.end());
  for (int i = 0; i < b.size(); ++i) {
    b[i].second = cnt[b[i].first];
    if (i) b[i].second += b[i - 1].second;
  }
  for (int i = 1; i <= n; ++i) {
    int p = bs(a[i].first);
    if (p < b.size()) ans[i] += b[p].second;
    if (ans[i] < 0) ans[i] = 0;
    cout << ans[i] << ' ';
  }
  return 0;
}
