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 const M = 1000000007;
double const pi = acos(-1);
long long const inf = 2e18;
long long const N = 200001;
vector<vector<long long> > g(N);
vector<long long> a(N), cnt(N, 0);
map<long long, vector<long long> > temp;
long long c = 0;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, t, n, k;
  cin >> n >> k;
  set<long long> s;
  map<long long, vector<long long> > m;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    m[a[i]].push_back(i);
  }
  for (i = 1; i <= k; i++) {
    long long f, s;
    cin >> f >> s;
    g[f].push_back(a[s]);
    g[s].push_back(a[f]);
  }
  for (i = 1; i <= n; i++) {
    g[i].push_back(0);
    sort(g[i].begin(), g[i].end());
    long long index =
        lower_bound(g[i].begin(), g[i].end(), a[i]) - g[i].begin();
    cnt[i] = index - 1;
  }
  vector<long long> ans(n + 1);
  long long counter = 0;
  for (auto itr = m.begin(); itr != m.end(); itr++) {
    for (auto x : itr->second) ans[x] = counter;
    counter += ((itr->second).size());
  }
  for (i = 1; i <= n; i++) cout << ans[i] - cnt[i] << " ";
  return 0;
}
