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 MAX = 2e5 + 1;
struct p {
  int a, id, ID, x;
  bool operator<(const p& b) const { return x > b.x; }
} c[MAX];
bool cmp(p a, p b) { return a.a < b.a; }
int n, num, m;
int ans[MAX], ID[MAX], sum[MAX], he[MAX];
int read() {
  int x = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    ;
  for (; isdigit(ch); x = x * 10 + ch - 48, ch = getchar())
    ;
  return x;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) c[i].a = read(), c[i].ID = i;
  sort(c + 1, c + 1 + n, cmp);
  for (int i = 1; i <= n; ++i) {
    if (c[i].a != c[i - 1].a) ++num;
    c[i].x = num, ++sum[num];
    ID[c[i].ID] = i;
  }
  for (int i = 1; i <= m; ++i) {
    int x = read(), y = read();
    if (c[ID[x]].x > c[ID[y]].x)
      ++c[ID[x]].id;
    else if (c[ID[x]].x < c[ID[y]].x)
      ++c[ID[y]].id;
  }
  sort(c + 1, c + 1 + n);
  int x = 1;
  for (int i = 1; i <= n; ++i) {
    if (c[i].x != c[i - 1].x) x = i;
    ans[c[i].ID] = n - x - sum[c[i].x] + 1 - c[i].id;
  }
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
