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.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Main().solve(in, out);
        out.close();
    }

    private void solve(FastScanner in, PrintWriter out) {
        int n = in.nextInt(), k = in.nextInt();
        Node[] g = new Node[n];
        for (int i = 0; i < n; ++i) {
            g[i] = new Node(i, in.nextInt());
        }
        for (int i = 0; i < k; ++i) {
            int u = in.nextInt() - 1, v = in.nextInt() - 1;
            g[u].addNeighbor(v);
            g[v].addNeighbor(u);
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            for (int u: g[i].neighbors) {
                if (g[u].skill < g[i].skill) {
                    --ans[i];
                }
            }
        }
        Arrays.sort(g);
        for (int i = 0; i < n; ) {
            int j = i + 1;
            while (j < n && g[j].skill == g[i].skill) {
                ++j;
            }
            for (int t = i; t < j; ++t) {
                ans[g[t].id] += i;
            }
            i = j;
        }
        for (int i = 0; i < n; ++i) {
            out.print(ans[i] + " ");
        }
        out.println();
    }

    class Node implements Comparable<Node> {
        int id, skill;
        ArrayList<Integer> neighbors = new ArrayList<>();

        Node(int id, int skill) {
            this.id = id;
            this.skill = skill;
        }

        void addNeighbor(int v) {
            neighbors.add(v);
        }

        @Override
        public int compareTo(Node o) {
            return skill - o.skill;
        }
    }

    private static class FastScanner {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        public FastScanner(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
        }

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

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

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

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