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 = 200005;
struct Node {
  int val;
  Node *ls, *rs;
  Node() {
    val = 0;
    ls = rs = NULL;
  }
} pool[MAXN << 5], *root[MAXN];
int ans[MAXN];
Node *newNode() {
  static int cnt = 0;
  if (cnt >= MAXN << 5) return new Node();
  return &pool[cnt++];
}
void insert(Node *&cur, int l, int r, int p, int v) {
  if (!cur) cur = newNode();
  cur->val += v;
  if (l < r) {
    int mid = (l + r) / 2;
    if (p <= mid)
      insert(cur->ls, l, mid, p, v);
    else
      insert(cur->rs, mid + 1, r, p, v);
  }
}
int query(Node *cur, int l, int r, int a, int b) {
  if (a > b) return 0;
  if (!cur) return 0;
  int res = 0;
  if (a <= l && b >= r) {
    res = cur->val;
  } else {
    int mid = (l + r) / 2;
    if (a <= mid) res += query(cur->ls, l, mid, a, b);
    if (b > mid) res += query(cur->rs, mid + 1, r, a, b);
  }
  return res;
}
int r[MAXN], pa[MAXN];
int find(int x) { return pa[x] == x ? pa[x] : pa[x] = find(pa[x]); }
void merge(int x, int y) {
  x = find(x), y = find(y);
  if (x != y) pa[x] = y;
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  vector<int> disp;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &r[i]);
    disp.push_back(r[i]);
  }
  sort(disp.begin(), disp.end());
  disp.erase(unique(disp.begin(), disp.end()), disp.end());
  for (int i = 1; i <= n; i++) {
    r[i] = lower_bound(disp.begin(), disp.end(), r[i]) - disp.begin() + 1;
  }
  for (int i = 1; i <= n; i++) pa[i] = i;
  for (int i = 1; i <= k; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (r[x] > r[y])
      ans[x]--;
    else if (r[x] < r[y])
      ans[y]--;
    merge(x, y);
  }
  for (int i = 1; i <= n; i++) {
    insert(root[0], 1, n, r[i], 1);
    insert(root[find(i)], 1, n, r[i], 1);
  }
  for (int i = 1; i <= n; i++) {
    ans[i] += query(root[0], 1, n, 1, r[i] - 1);
    printf("%d ", ans[i]);
  }
  putchar('\n');
  return 0;
}
