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.util.*;
import java.io.*;

public class Task{

 static class pair{
	long c;int idx;
	pair(long c, int idx){
		this.c=c;
		this.idx=idx;
	}
 }
static BufferedReader s1;
static BufferedWriter out;
static String read() throws IOException{String line="";while(line.length()==0){line=s1.readLine();continue;}return line;}
static int int_v (String s1){return Integer.parseInt(s1);}
static long long_v(String s1){return Long.parseLong(s1);}
static int[] int_arr() throws IOException{String[] a=read().split(" ");int[] b=new int[a.length];for(int i=0;i<a.length;i++){b[i]=int_v(a[i]);}return b;}
static long[] long_arr() throws IOException{String[] a=read().split(" ");long[] b=new long[a.length];for(int i=0;i<a.length;i++){b[i]=long_v(a[i]);}return b;}
static void assign(){s1=new BufferedReader(new InputStreamReader(System.in));out=new BufferedWriter(new OutputStreamWriter(System.out));}
//......................................@uthor_Alx..............................................
static int f(int[] a, int[] b) {
	return a[0]-b[0];
}
static int bs(int[][] a, int v){
	int l=0,r=a.length-1;
	while(l<=r){
		int m=(l+r)/2;
		if(a[m][0]>=v){
			r=m-1;
		}
		else{
			l=m+1;
		}
	}
	return l;
}
public static void main(String[] args) throws  IOException{
				assign();
				int t=1;//int_v(read()),cn=1;
				while(t--!=0){
					int[] z=int_arr();
					int n=z[0],k=z[1];
					int[] arr=int_arr();
					int[] d=new int[n+1];
					for(int i=0;i<k;i++){
						int[] x=int_arr();
						if(arr[x[0]-1]>arr[x[1]-1]) d[x[0]]--;
						else if(arr[x[1]-1]>arr[x[0]-1])d[x[1]]--;
					}
					int[][] a=new int[n][2];
					for(int i=0;i<n;i++){
						a[i][0]=arr[i];a[i][1]=i+1;
					}
					Arrays.sort(a,(x,y)->f(x,y));
					int[] res=new int[n];
					for(int i=0;i<n;i++){
						res[a[i][1]-1]=Math.max(0,bs(a,a[i][0])+d[a[i][1]]);
						//out.write(d[i+1]+" "+bs(a,a[i][0])+" \n");
					}
					for(int i=0;i<n;i++){
						out.write(res[i]+" ");
					}
				}			    
			    out.flush();

	}
}