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 mod = 1e9 + 7;
const long long mod1 = 998244353;
const double pie = 3.1415926535;
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else {
    if (y % 2 == 0)
      return power(x * x, y / 2);
    else
      return x * power(x * x, (y - 1) / 2);
  }
}
long long mod_power(long long x, long long y, long long m) {
  long long r = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) r = (r * x) % m;
    y = y / 2;
    x = (x * x) % m;
  }
  return r;
}
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
vector<long long> extended_Euclid(long long a, long long b) {
  vector<long long> v(3);
  if (b == 0) {
    v[0] = a;
    v[1] = 1;
    v[2] = 0;
    return v;
  } else {
    vector<long long> v1 = extended_Euclid(b, a % b);
    v[0] = v1[0];
    v[1] = v1[2];
    v[2] = v1[1] - (a / b) * v1[2];
    return v;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int TESTS = 1;
  while (TESTS--) {
    int n, k;
    cin >> n >> k;
    vector<pair<int, int> > v;
    vector<int> arr(n + 1);
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      arr[i] = x;
      v.push_back(make_pair(x, i));
    }
    sort(v.begin(), v.end());
    map<int, int> M;
    map<int, int> adj;
    for (int i = 0; i < k; i++) {
      int x, y;
      cin >> x >> y;
      if (arr[x] == arr[y])
        continue;
      else {
        if (arr[x] > arr[y])
          adj[x]++;
        else
          adj[y]++;
      }
    }
    int flag = 1;
    for (int i = 0; i < n; i++) {
      if (i == 0) {
        M[v[i].second] = 0;
      } else {
        if (v[i].first == v[i - 1].first) {
          M[v[i].second] = i - flag - adj[v[i].second];
          flag++;
        } else {
          M[v[i].second] = i - adj[v[i].second];
          flag = 1;
        }
      }
    }
    auto ip = M.begin();
    for (ip; ip != M.end(); ip++) cout << ip->second << " ";
    cout << endl;
  }
}
