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;
void merge(long long int arr[], long long int l, long long int m,
           long long int r) {
  long long int i, j, k;
  long long int n1 = m - l + 1;
  long long int n2 = r - m;
  long long int L[n1], R[n2];
  for (i = 0; i < n1; i++) L[i] = arr[l + i];
  for (j = 0; j < n2; j++) R[j] = arr[m + 1 + j];
  i = 0;
  j = 0;
  k = l;
  while (i < n1 && j < n2) {
    if (L[i] <= R[j]) {
      arr[k] = L[i];
      i++;
    } else {
      arr[k] = R[j];
      j++;
    }
    k++;
  }
  while (i < n1) {
    arr[k] = L[i];
    i++;
    k++;
  }
  while (j < n2) {
    arr[k] = R[j];
    j++;
    k++;
  }
}
void mergeSort(long long int arr[], long long int l, long long int r) {
  if (l < r) {
    long long int m = l + (r - l) / 2;
    mergeSort(arr, l, m);
    mergeSort(arr, m + 1, r);
    merge(arr, l, m, r);
  }
}
long long int binarySearch(long long int arr[], long long int l,
                           long long int r, long long int x) {
  while (l <= r) {
    long long int m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
long long int ans(long long int t, long long int a[]) {
  long long int temp = 0;
  for (long long int i = t; i >= 0; i--) {
    if (a[i] == a[t]) {
      continue;
    } else {
      temp = i + 1;
      break;
    }
  }
  return temp;
}
int main() {
  long long int n, k, temp1, temp2, temp;
  cin >> n >> k;
  long long int a[n], c[n], key[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    c[i] = a[i];
    key[i] = 0;
  }
  for (int i = 0; i < k; i++) {
    temp = 0;
    cin >> temp1;
    cin >> temp2;
    if (a[temp2 - 1] < a[temp1 - 1]) {
      key[temp1 - 1]--;
    }
    if (a[temp1 - 1] < a[temp2 - 1]) {
      key[temp2 - 1]--;
    }
  }
  sort(c, c + n);
  temp = 0;
  for (int i = 0; i < n; i++) {
    temp = std::lower_bound(c, c + n, a[i]) - c;
    key[i] += temp;
    cout << key[i] << " ";
  }
  return 0;
}
