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;
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, map<L, R> M) {
  os << "[";
  for (auto pr : M) os << pr.first << "->" << pr.second << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, unordered_map<L, R> M) {
  os << "[";
  for (auto pr : M) os << pr.first << "->" << pr.second << ",";
  return os << "]";
}
mt19937_64 mt(time(0));
struct Matrix {
  vector<vector<int>> data;
  int r, c;
  Matrix(int row, int col, bool identity = false) : r(row), c(col) {
    data.assign(row, vector<int>(col, 0));
    if (identity) {
      for (int i = 0; i < r; ++i) {
        data[i][i] = 1;
      }
    }
  }
  Matrix operator*(Matrix &other) {
    int m = r, n = c, p = other.c;
    Matrix res(m, p);
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < p; ++j) {
        for (int k = 0; k < n; ++k) {
          res.data[i][j] += data[i][k] * other.data[k][j];
        }
      }
    }
    return res;
  }
};
vector<int> extendGcd(int a, int b) {
  if (b == 0) {
    return {1, 0, a};
  } else {
    vector<int> tmp = extendGcd(b, a % b);
    return {tmp[1], tmp[0] - (a / b) * tmp[1], tmp[2]};
  }
}
Matrix matrix_power(Matrix base, long long exp) {
  int n = base.r;
  Matrix res(n, n, true);
  while (exp) {
    if (exp & 1) {
      res = res * base;
    }
    base = base * base;
    exp >>= 1;
  }
  return res;
}
int n, k;
int r[200010], res[200010];
vector<int> foo;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> r[i];
    foo.push_back(r[i]);
  }
  sort((foo).begin(), (foo).end());
  for (int i = 1; i <= n; ++i) {
    res[i] = lower_bound((foo).begin(), (foo).end(), r[i]) - foo.begin();
  }
  int u, v;
  for (int i = 1; i <= k; ++i) {
    cin >> u >> v;
    if (r[u] > r[v]) {
      res[u]--;
    } else if (r[u] < r[v]) {
      res[v]--;
    }
  }
  for (int i = 1; i <= n; ++i) {
    cout << res[i] << " \n"[i == n];
  }
  return 0;
}
