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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

    private FastScanner scanner = new FastScanner();

    public static void main(String[] args) {
        new Main().solve();
    }

    List<List<Integer>> gr = new ArrayList<>();

    boolean used[];
    List<Integer> ans = new ArrayList<>();


    private void solve() {
        int n = scanner.nextInt(), k = scanner.nextInt();
        int a[][] = new int[n][3], b[] = new int[n];

        List<List<Integer>> xy = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            a[i][0] = scanner.nextInt();
            b[i] = a[i][0];
            a[i][1] = i;
            xy.add(new ArrayList<>());
        }

        for (int i = 0; i < k; i++) {
            int x = scanner.nextInt() - 1 , y = scanner.nextInt() - 1;
            xy.get(x).add(y);
            xy.get(y).add(x);
        }

        Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);

        Map<Integer, Integer> used = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int cur = a[i][0];

            int t = i - used.getOrDefault(cur, 0);

            for (int y: xy.get(a[i][1])) {
                if (cur > b[y] ) {
                    t --;
                }
            }
            a[i][2] = t;
            used.put(cur, used.getOrDefault(cur,0) + 1);
        }

        Arrays.sort(a, (o1, o2) -> o1[1] - o2[1]);

        for (int i = 0; i < n; i++) {
            System.out.print(a[i][2] + " ");
        }
    }

    public static final <T> void swap(T[] a, int i, int j) {
        T t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    class Pair implements Comparable<Pair> {
        int a, i;

        public Pair(int a, int i) {
            this.a = a;
            this.i = i;
        }

        @Override
        public int compareTo(Pair o) {
            return this.a - o.a;
        }
    }

    long gcd(long a, long b) {
        if (b != 0) {
            return gcd(b, a % b);
        }
        return a;
    }

    void initGr(int n, int m) {
        for (int i = 0; i < n; i++) {
            gr.add(new ArrayList<>());
        }

        for (int i = 0; i < m; i++) {
            int u = scanner.nextInt() - 1, v = scanner.nextInt() - 1;
            gr.get(u).add(v);
            gr.get(v).add(u);
        }
    }

    class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
    }
}