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;
const long long int MOD = 1e9 + 7;
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
long long int binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  if (b % 2)
    return res * res * a;
  else
    return res * res;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<int> > r(n, vector<int>(2));
  for (int i = 0; i < n; i++) {
    cin >> r[i][0];
    r[i][1] = i;
  }
  vector<int> rem(n);
  for (int i = 0; i < k; i++) {
    int x, y;
    cin >> x >> y;
    if (r[x - 1][0] < r[y - 1][0]) {
      rem[y - 1] += 1;
    } else if (r[x - 1][0] > r[y - 1][0]) {
      rem[x - 1] += 1;
    }
  }
  vector<int> ans(n);
  sort(r.begin(), r.end());
  for (int i = 0; i < n; i++) {
    int x = -1;
    for (int b = n; b >= 1; b /= 2) {
      while (x + b < i && r[x + b][0] < r[i][0]) x += b;
    }
    ans[r[i][1]] = max(0, (x + 1) - rem[r[i][1]]);
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  for (int testCase = 1; testCase <= T; testCase++) {
    solve();
  }
}
