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 com.sun.source.tree.Tree;

import javax.naming.spi.DirObjectFactory;
import javax.xml.crypto.dsig.spec.XSLTTransformParameterSpec;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;


import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.nio.file.LinkOption;
import java.util.*;


public class Main {

   static boolean vis[][];
   static boolean vis2[][];
   static char[][]a;
   static boolean reach(int row,int col,int type){
        if (row>=vis.length||col>=vis.length)return false;
        if (a[row][col]=='F')return true;
        if (vis[row][col]||(a[row][col]!='S'&&(a[row][col]-'0')!=type))return false;
        vis[row][col]=true;
        boolean state=false;
       state|= reach(row+1,col,type);
        state|=reach (row,col+1,type);
        return state;
    }

    static boolean reach2(int row,int col,int type){
        if (row>=vis2.length||col>=vis2.length)return false;
        if (a[row][col]=='F')return true;

        if (vis2[row][col]||(a[row][col]!='S'&&(a[row][col]-'0')!=type))return false;
        vis2[row][col]=true;
        boolean state = reach2(row + 1, col, type);
        state|=reach2(row,col+1,type);
        return state;
    }



    public static void main(String[] args) throws IOException {
        Scanner in = new Scanner(System.in);


        try (PrintWriter or = new PrintWriter(System.out)) {

            int t = 1;
            outer:
            while (t-- > 0) {


               int n = in.nextInt(),k=in.nextInt();




                int [][]a=new int[n][2];
                for (int i = 0; i < n; i++) {
                    a[i][0]=in.nextInt();
                    a[i][1]=i;
                }

                int []ans=new int[n];


                for (int i = 0; i < k; i++) {
                    int x=in.nextInt()-1;
                    int y =in.nextInt()-1;
                    if (a[y][0]<a[x][0])ans[a[x][1]]--;
                    else if (a[y][0]>a[x][0])ans[a[y][1]]--;
                }
                Arrays.sort(a, new Comparator<int[]>() {
                    @Override
                    public int compare(int[] o1, int[] o2) {
                        return o1[0]-o2[0];
                    }
                });

                //(int y:ans)or.print(y+" ");
                or.println();
                for (int i = 0; i < n; i++) {
                    int start=0,end=n-1,tt=-1;
                    while (start<=end){
                        int mid=(start+end)/2;
                        if (a[mid][0]>=a[i][0])end=mid-1;
                        else {
                            tt=mid;
                            start=mid+1;
                        }
                    }
                    if (tt!=-1) ans[a[i][1]]+=tt+1;
                    ans[a[i][1]]=Math.max(ans[a[i][1]],0);
                }
                for(int y:ans)or.print(y+" ");












            }
        }


    }

    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

}


