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 maxn = 2e6 + 10;
int n, ans[maxn];
struct point {
  long long id;
  long long v;
} pt[maxn], pp[maxn];
bool cmp(point a, point b) { return a.v < b.v; }
bool cmp1(point a, point b) { return a.id < b.id; }
int s_s(long long v) {
  int l = 0, r = n - 1, po, mid;
  while (l < r) {
    mid = (l + r) >> 1;
    if (pt[mid].v >= v)
      r = mid;
    else
      l = mid + 1;
  }
  return r;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long sum = 0, po, poo, a, b, k;
  cin >> n >> k;
  memset(ans, 0, sizeof(ans));
  for (long long i = 0; i < n; i++) {
    cin >> pt[i].v;
    pt[i].id = i;
  }
  for (long long i = 0; i < k; i++) {
    cin >> a >> b;
    a -= 1;
    b -= 1;
    if (pt[a].v > pt[b].v)
      ans[a]++;
    else if (pt[b].v > pt[a].v)
      ans[b]++;
  }
  int q = 0;
  sort(pt, pt + n, cmp);
  for (long long i = 0; i < n; i++) {
    sum = 0;
    po = pt[i].id;
    poo = s_s(pt[i].v);
    if (poo != i)
      sum = (i - ans[po] - (i - poo));
    else
      sum = i - ans[po];
    pp[q].v = sum;
    pp[q++].id = pt[i].id;
  }
  sort(pp, pp + q, cmp1);
  for (int i = 0; i < q; i++) cout << pp[i].v << " ";
  cout << endl;
  return 0;
}
