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;
bool prime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool subsequence_checker(string str1, string str2, long long int m,
                         long long int n) {
  if (m == 0) return true;
  if (n == 0) return false;
  if (str1[m - 1] == str2[n - 1])
    return subsequence_checker(str1, str2, m - 1, n - 1);
  return subsequence_checker(str1, str2, m, n - 1);
}
long long int inv(long long int a) {
  long long int p = 998244353;
  long long int b = p - 2, ans = 1;
  for (; b > 0; a = a * a % p, b >>= 1)
    if (b & 1) ans = ans * a % p;
  return ans;
}
long long int ncr(long long int n, long long int m) {
  long long int p = 998244353;
  if (m > n) return 0;
  long long int up = 1, down = 1;
  for (int i = n - m + 1; i < n + 1; i++) up *= i, up %= p;
  for (int i = 1; i < m + 1; i++) down *= i, down %= p;
  return up * inv(down) % p;
}
long long int countDigit(long long int n) { return floor(log10(n) + 1); }
long long int a_pow_b(long long int a, long long int b) {
  long long int p = 1000000007;
  long long int ans = 1;
  for (; b > 0; a = a * a % p, b >>= 1)
    if (b & 1) ans = ans * a % p;
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
bool isPerfectSquare(long long int x) {
  long double sr = sqrt(x);
  return ((sr - floor(sr)) == 0);
}
long long int ncr123(long long int n, long long int k) {
  if (k == 1) {
    return n;
  }
  if (k == 2) {
    return n * (n - 1) / 2;
  }
  if (k == 3) {
    return n * (n - 1) * (n - 2) / 6;
  }
}
bool comp(const pair<long long int, long long int> &a,
          const pair<long long int, long long int> &b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second < b.second;
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> v;
  for (long long int i = 0; i < n; i++) {
    long long int var;
    cin >> var;
    v.push_back(var);
  };
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) a[i] = v[i];
  map<long long int, long long int> m;
  while (k--) {
    long long int s, e;
    cin >> s >> e;
    s--;
    e--;
    if (a[s] > a[e]) {
      m[s]++;
    } else if (a[s] < a[e]) {
      m[e]++;
    }
  }
  sort(a.begin(), a.end());
  for (long long int i = 0; i < n; i++) {
    auto pos = lower_bound(a.begin(), a.end(), v[i]) - a.begin();
    long long int ans = pos - m[i];
    if (ans <= 0) {
      cout << 0 << " ";
    } else
      cout << ans << " ";
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t;
  t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
