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.*;

public class cf {

	static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e){e.printStackTrace();}
        }

        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {st = new StringTokenizer(br.readLine());}
            catch (Exception e) {e.printStackTrace();}
            return st.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() {
            String line = "";
            if(st.hasMoreTokens()) line = st.nextToken();
            else try {return br.readLine();}catch(IOException e){e.printStackTrace();}
            while(st.hasMoreTokens()) line += " "+st.nextToken();
            return line;
        }
    }
    
    static class Pair {
        int val;
        int cnt;
        int idx;
        public Pair(int val, int cnt, int idx) {
            this.val = val;
            this.cnt = cnt;
            this.idx = idx;
        }
    }
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter pw = new PrintWriter(System.out);

        int n = sc.nextInt();
        int k = sc.nextInt();
        Pair[] pairs = new Pair[n];

        for(int i=0;i<n;i++) pairs[i] = new Pair(sc.nextInt(), 0, i);

        for(int i=0;i<k;i++) {
            int x = sc.nextInt() - 1;
            int y = sc.nextInt() - 1;
            if(pairs[x].val > pairs[y].val) pairs[x].cnt++;
            if(pairs[y].val > pairs[x].val) pairs[y].cnt++;
        }

        Arrays.sort(pairs, (Pair cur, Pair other) -> (cur.val - other.val));

        /*pw.print("val: ");
        for(Pair p : pairs) pw.print(p.val + " ");
        pw.println();
        pw.print("idx: ");
        for(Pair p : pairs) pw.print(p.idx + " ");
        pw.println();
        pw.print("cnt: ");
        for(Pair p : pairs) pw.print(p.cnt + " ");
        pw.println();*/

        int[] ans = new int[n];
        for(int i=0;i<n;i++) {
            int lo = 0;
            int hi = n - 1;
            int res = 0;
            while(lo <= hi) {
                int mid = lo + (hi - lo) / 2;
                if(pairs[mid].val == pairs[i].val) {
                    res = mid;
                    hi = mid - 1;
                }
                else if(pairs[mid].val > pairs[i].val) {
                    hi = mid - 1;
                }
                else lo = mid + 1;
            }
            ans[pairs[i].idx] = res - pairs[i].cnt;
        }

        for(int res : ans) pw.print(res + " ");
        pw.close();
	}

}