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;
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &v) {
  os << "(";
  os << v.first << "," << v.second << ")";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  long long sz = v.size();
  for (long long i = 0; i < sz; ++i) {
    os << v[i];
    if (i != sz - 1) os << ",";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  T last = *v.rbegin();
  os << "[";
  for (auto it : v) {
    os << it;
    if (it != last) os << ",";
  }
  os << "]\n";
  return os;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, const map<T, S> &v) {
  for (auto it : v) {
    os << it.first << " : " << it.second << "\n";
  }
  return os;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 1000000007;
    }
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return res;
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long N = 2e5 + 5;
long long r[N];
long long val[N];
vector<long long> arr[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (__typeof(1) i = 1; i <= n; i++) cin >> r[i];
  for (__typeof(1) i = 1; i <= n; i++) val[i] = r[i];
  for (__typeof(1) i = 1; i <= k; i++) {
    long long x, y;
    cin >> x >> y;
    arr[x].push_back(r[y]);
    arr[y].push_back(r[x]);
  }
  for (__typeof(1) i = 1; i <= n; i++) sort((arr[i]).begin(), (arr[i]).end());
  sort(r + 1, r + n + 1);
  for (__typeof(1) i = 1; i <= n; i++) {
    long long low = 1, high = n, mid, res = 0;
    while (low <= high) {
      mid = (low + high) >> 1;
      if (r[mid] < val[i]) {
        res = mid;
        low = mid + 1;
      } else
        high = mid - 1;
    }
    long long ans = res;
    long long sz = arr[i].size();
    low = 0, high = sz - 1, res = -1;
    while (low <= high) {
      mid = (low + high) >> 1;
      if (arr[i][mid] < val[i]) {
        res = mid;
        low = mid + 1;
      } else
        high = mid - 1;
    }
    ans -= (res + 1);
    cout << ans << " ";
  }
  cout << endl;
  return 0;
}
