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 C{
	public static void main(String[] args)throws Throwable {
		MyScanner sc=new MyScanner();
		PrintWriter pw=new PrintWriter(System.out);
		
		int n=sc.nextInt();
		int m=sc.nextInt();
		ArrayList<Integer> [] adj=new ArrayList [n];
		for(int i=0;i<n;i++)
			adj[i]=new ArrayList<>();
		pro [] a=new pro[n];
		for(int i=0;i<n;i++)
			a[i]=new pro(i, sc.nextInt());
		while(m-->0) {
			int x=sc.nextInt()-1;
			int y=sc.nextInt()-1;
			if(a[x].r<a[y].r)
				adj[x].add(y);
			else
				if(a[y].r<a[x].r)
					adj[y].add(x);
		}
		Arrays.sort(a);
		int [] ans=new int [n];
		int cntSame=1;
		for(int i=0;i<n;i++){
			if(i>0 && a[i].r==a[i-1].r)
				cntSame++;
			else
				cntSame=1;
			for(int x : adj[a[i].idx])
				ans[x]--;
			ans[a[i].idx]+=(i-(cntSame-1));
		}
		
		for(int x : ans)
			pw.print(x+" ");
		pw.flush();
		pw.close();
    }
	
	static class pro implements Comparable<pro>{
		int idx,c,r;
		
		pro(int i,int r){
			c=0;
			this.r=r;
			idx=i;
		}
		public int compareTo(pro a) {
			return r-a.r;
		}
	}
	
	
	static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {while (st == null || !st.hasMoreElements()) {
			try {st = new StringTokenizer(br.readLine());}
			catch (IOException e) {e.printStackTrace();}}
		return st.nextToken();}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		String nextLine(){String str = "";
		try {str = br.readLine();}
		catch (IOException e) {e.printStackTrace();}
		return str;}
	}
} 