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;
struct Node {
  int skills, num;
} a[200005], b[200005];
bool cmp(Node x, Node y) {
  if (x.skills != y.skills)
    return x.skills < y.skills;
  else
    return x.num < y.num;
}
int dir[200005];
int main() {
  int n, k, x, y, i, j;
  int t, first, last, mid, ans;
  memset(dir, 0, sizeof(dir));
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i].skills);
    a[i].num = i;
    b[i] = a[i];
  }
  for (i = 0; i < k; i++) {
    scanf("%d%d", &x, &y);
    x--;
    y--;
    if (a[x].skills < a[y].skills)
      dir[y]++;
    else if (a[x].skills > a[y].skills)
      dir[x]++;
  }
  sort(b, b + n, cmp);
  for (i = 0; i < n; i++) {
    t = a[i].skills;
    first = 0;
    last = n - 1;
    mid = 0;
    while (first <= last) {
      mid = (first + last) / 2;
      if (b[mid].skills < t)
        first = mid + 1;
      else if (b[mid].skills >= t)
        last = mid - 1;
    }
    ans = (last + 1) - dir[a[i].num];
    printf("%d ", ans);
  }
  return 0;
}
