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 int md = 100100;
int n, m, x, y;
int ans[200100];
int a[200100];
pair<int, int> ps[200100];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> ps[i].first;
    a[i] = ps[i].first;
    ps[i].second = i;
  }
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    if (a[x] < a[y])
      --ans[y];
    else if (a[y] < a[x])
      --ans[x];
  }
  sort(ps + 1, ps + n + 1);
  int k = 0;
  for (int i = 1; i <= n; i++) {
    int id = ps[i].second;
    int vl = ps[i].first;
    while (k + 1 <= n && ps[k + 1].first < vl) ++k;
    ans[id] += k;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
