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 N = 200005;
int n, m, pos[N], ans[N];
vector<int> g[N];
map<int, int> tot;
struct data {
  int a, i;
} a[N];
bool cmp(data u, data v) { return u.a > v.a; }
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i].a), a[i].i = i;
    ++tot[a[i].a];
  }
  sort(a + 1, a + 1 + n, cmp);
  for (int i = 1; i <= n; ++i) pos[a[i].i] = i;
  for (int i = 1, x, y; i <= m; ++i) {
    scanf("%d%d", &x, &y);
    x = pos[x], y = pos[y];
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int cnt = n;
  a[0].a = -1;
  for (int i = 1; i <= n; ++i) {
    int s = g[i].size();
    if (a[i].a != a[i - 1].a) cnt -= tot[a[i].a];
    ans[a[i].i] = cnt;
    for (int j = 0; j < s; ++j)
      if (g[i][j] > i && a[g[i][j]].a != a[i].a) --ans[a[i].i];
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
