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;
inline int getint() {
  char c = getchar();
  int x = 0;
  int minus = 1;
  while (c != EOF && c != '-' && !isdigit(c)) c = getchar();
  if (c == '-') {
    minus = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * minus;
}
inline long long getll() {
  long long x;
  scanf("%lld", &x);
  return x;
}
double getdb() {
  static double x;
  scanf("%lf", &x);
  return x;
}
template <typename T>
void updmax(T& a, T const& b) {
  a = max(a, b);
}
template <typename T>
void updmin(T& a, T const& b) {
  a = min(a, b);
}
template <typename T>
int Descr(T* from, T* to) {
  sort(from, to);
  return (int)(unique(from, to) - from);
}
template <typename T>
int LBound(T* from, T* to, T const& val) {
  return (int)(lower_bound(from, to, val) - from);
}
template <typename T>
int UBound(T* from, T* to, T const& val) {
  return (int)(upper_bound(from, to, val) - from);
}
const int mxn = 2e5 + 50;
int sk[mxn];
int s[mxn];
int sp[mxn];
int ans[mxn];
int main() {
  int n = getint();
  int m = getint();
  for (int i = 1, lim__lim = (n); i <= lim__lim; i++) sk[i] = getint();
  for (int j = 1, lim__lim = (m); j <= lim__lim; j++) {
    int a = getint();
    int b = getint();
    if (sk[a] > sk[b]) swap(a, b);
    if (sk[a] != sk[b]) sp[b]++;
  }
  for (int i = 1, lim__lim = (n); i <= lim__lim; i++) s[i] = i;
  sort(s + 1, s + n + 1,
       [](int const& a, int const& b) { return sk[a] < sk[b]; });
  int cx = 0;
  for (int i = 1, lim__lim = (n); i <= lim__lim; i++) {
    while (cx <= n && sk[s[cx]] < sk[s[i]]) cx++;
    ans[s[i]] = cx - 1 - sp[s[i]];
  }
  for (int i = 1, lim__lim = (n); i <= lim__lim; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}
