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;
struct gg {
  int w, id;
} node[200010];
int n, k, r[200010], ans[200010], nw[200010], id[200010];
vector<int> mp[200010];
bool cmp(gg mo, gg ha) { return mo.w < ha.w; }
int find(int l, int r, int x, int w) {
  int ans = -1, mid;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (mp[x][mid] < w)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> node[i].w, node[i].id = i;
  int u, v;
  for (int i = 1; i <= k; i++)
    cin >> u >> v, mp[u].push_back(node[v].w), mp[v].push_back(node[u].w);
  for (int i = 1; i <= n; i++) sort(mp[i].begin(), mp[i].end());
  sort(node + 1, node + n + 1, cmp);
  for (int i = 1; i <= n; i++) nw[i] = node[i].w, id[i] = node[i].id;
  int pos, ps;
  for (int i = 1; i <= n; i++) {
    pos = lower_bound(nw + 1, nw + n + 1, nw[i]) - nw;
    if (pos > n) pos = n;
    while (nw[pos] == nw[i] && pos >= 1) pos--;
    ps = find(0, mp[id[i]].size() - 1, id[i], nw[i]) + 1;
    ans[id[i]] = max(0, pos - ps);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
