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 long long MOD = 1e9 + 7;
const double PI = acos(-1.0);
const double EPS = (1e-10);
const int dx4[4] = {0, 1, 0, -1};
const int dy4[4] = {-1, 0, 1, 0};
const int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
bool cmp(int X, int Y) { return X > Y; }
int dcp(double A, double B) { return fabs(A - B) <= EPS ? 0 : A < B ? -1 : 1; }
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long modpow(long long a, long long b, long long m = MOD) {
  a %= m;
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long bpow(long long a, long long b) {
  if (b < 0) {
    return 0;
  }
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a, long long m = MOD) { return modpow(a, m - 2, m); }
const int N = 1e6 + 5;
vector<int> adj[N];
bool visited[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  set<long long> s;
  map<long long, long long> cnt, fre;
  long long a[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    fre[a[i]]++;
    s.insert(a[i]);
  }
  long long c = 0;
  for (auto x : s) {
    cnt[x] = c;
    c = c + fre[x];
  }
  long long ans[n + 1];
  for (int i = 1; i <= n; i++) {
    ans[i] = cnt[a[i]];
  }
  for (int i = 0; i < k; i++) {
    long long x, y;
    cin >> x >> y;
    if (a[x] > a[y]) {
      ans[x]--;
    } else if (a[x] < a[y]) {
      ans[y]--;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}
