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.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;


public class vk18
{
    public static void main(String[] stp) throws Exception
    {
        Scanner scan = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken()),i,j,k=Integer.parseInt(st.nextToken());
        Long input[]=new Long[n];
        st = new StringTokenizer(br.readLine());
        TreeMap<Long,Integer> tm=new TreeMap<>();
        int total=0;
        for(i=0;i<n;i++)
        {
            input[i]=Long.parseLong(st.nextToken());
            int count=tm.getOrDefault(input[i],0);
            if(count==0) total++;
            tm.put(input[i],count+1);
        }
        Long a[]=new Long[total];
        Integer b[]=new Integer[total];
        for(i=0;i<total;i++)
        {
            a[i]=tm.firstKey();
            b[i]=tm.get(a[i]);
            tm.remove(a[i]);
        }
        //for(i=0;i<total;i++) pw.println(a[i]+" "+b[i]);
        Integer c[]=new Integer[total]; c[0]=0;
        for(i=1;i<total;i++) c[i]=c[i-1]+b[i-1];
        b[0]=0;
        int index1=0,index2=0;
        //for(i=0;i<total;i++) pw.print(c[i]+" ");
        //TreeSet<Long> ts=new TreeSet<>();
        Integer occ[]=new Integer[n];
        for(i=0;i<n;i++)
        {
            int index=Arrays.binarySearch(a,input[i]);
            occ[i]=c[index];
        }
        long first=0,second=0;
        for(i=0;i<k;i++)
        {
            st = new StringTokenizer(br.readLine());
            index1=Integer.parseInt(st.nextToken())-1;
            index2=Integer.parseInt(st.nextToken())-1;
            first=input[index1];
            second=input[index2];
            if(first > second) occ[index1]--;
            else if(second > first) occ[index2]--;
        }
        for(i=0;i<n;i++) pw.print(occ[i]+" ");
        pw.flush();
    }
}