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;
long long mod = 1000000007;
long long int lowerBound(long long int a[], long long int n, long long int x) {
  long long int l = 0, r = n - 1;
  long long int ans = -1;
  long long int mid = (l + r) / 2;
  while (l < r) {
    mid = (l + r) / 2;
    if (a[mid] == x) {
      if (a[mid - 1] != x) {
        r = mid;
        break;
      } else
        r = mid - 1;
    } else if (a[mid] < x) {
      l = mid + 1;
    } else
      r = mid - 1;
  }
  if (a[r] != x) return -1;
  return r;
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  long long int a[n];
  vector<long long int> c;
  vector<pair<long long int, long long int> > v;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    c.push_back(a[i]);
    v.push_back(make_pair(a[i], i));
  }
  long long int b[n];
  sort(v.begin(), v.end());
  sort(c.begin(), c.end());
  for (long long int i = 0; i < n; i++) {
    long long int j = v[i].second;
    long long int val = v[i].first;
    long long int lb = lower_bound(c.begin(), c.end(), val) - c.begin();
    b[j] = min(i, lb);
  }
  for (long long int i = 0; i < k; i++) {
    long long int x, y;
    cin >> x >> y;
    long long int n1 = a[x - 1];
    long long int n2 = a[y - 1];
    if (n1 > n2) {
      b[x - 1]--;
    } else if (n2 > n1)
      b[y - 1]--;
  }
  for (long long int i = 0; i < n; i++) cout << b[i] << " ";
  cout << '\n';
  return 0;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int fast_power(long long int base, long long int power,
                         long long int MOD) {
  long long int result = 1;
  while (power > 0) {
    if (power & 1) result = (result * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return result;
}
