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<long long, pair<int, int>>> v;
vector<int> par(1000000);
vector<int> sz(1000000);
void print(vector<int> v) {
  vector<int>::iterator it = v.begin();
  while (it != v.end()) {
    cout << *it << " ";
    it++;
  }
  cout << endl;
}
void create(int n) {
  for (int i = 1; i <= n; i++) {
    par[i] = i;
    sz[i] = 1;
  }
}
int fp(int n) {
  if (par[n] == n) return n;
  return par[n] = fp(par[n]);
}
void merg(int x, int y) {
  x = fp(x);
  y = fp(y);
  if (sz[x] < sz[y]) {
    swap(x, y);
  }
  par[y] = x;
  sz[x] += sz[y];
}
bool check(string s) {
  int n = s.size();
  int i = 0;
  int j = n - 1;
  bool deb = 1;
  while (i <= j) {
    if (s[i] != s[j]) {
      deb = 0;
      break;
    }
    i++;
    j--;
  }
  return deb;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> v(n + 1);
  vector<pair<int, int>> u;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    u.push_back(make_pair(v[i], i));
  }
  sort(u.begin(), u.end());
  vector<vector<int>> g(n + 1);
  for (int i = 0; i < k; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  vector<int> ans(n + 1);
  int prev = -1;
  ans[u[0].second] = 0;
  for (int i = 1; i < n; i++) {
    if (u[i].first == u[i - 1].first && prev == -1) {
      ans[u[i].second] = 0;
    } else {
      if (u[i].first != u[i - 1].first) {
        prev = i - 1;
      }
      int a = prev + 1;
      int pn = u[i].second;
      for (int j = 0; j < g[pn].size(); j++) {
        int nd = g[pn][j];
        if (v[nd] < v[pn]) a--;
      }
      ans[u[i].second] = a;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
