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>
#pragma GCC target("avx,avx2")
#pragma GCC optimize("Ofast,unroll-loops")
using namespace std;
using UL = unsigned long int;
using UI = unsigned int;
using ULL = unsigned long long int;
using LI = long int;
using LL = long long;
using LF = double;
using LLF = long double;
const int mod1 = 1000000007;
const int mod2 = 998244353;
const double PI = acos(-1);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int ntc = 1, tc, i, j;
  LL n, w, mn, mx;
  for (tc = 1; tc <= ntc; tc++) {
    cin >> n >> w;
    vector<LL> A(n);
    vector<LL> B(n);
    for (i = 0; i <= n - 1; i++) {
      cin >> A[i];
      B[i] = A[i];
    }
    sort(A.begin(), A.end());
    vector<int> ans(n, 0);
    for (i = 0; i <= n - 1; i++) {
      ans[i] = (lower_bound(A.begin(), A.end(), B[i]) - A.begin());
    }
    LL a, b;
    while (w--) {
      cin >> a >> b;
      a--;
      b--;
      if (B[a] < B[b]) {
        ans[b]--;
      }
      if (B[a] > B[b]) ans[a]--;
    }
    for (i = 0; i <= n - 1; i++) {
      cout << ans[i] << " ";
    }
  }
  return 0;
}
