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 C481F
{
	public static void main(String[] args)
	{
		FastScanner in = new FastScanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		Programmer[] arr = new Programmer[n];
		for (int i = 0; i < n; i++)
			arr[i] = new Programmer(i, in.nextInt());
		Arrays.sort(arr);
		int[] convert = new int[n];
		for (int i = 0; i < n; i++)
			convert[arr[i].i] = i;
		
		int[] ans = new int[n];
		for (int i = 1; i < n; i++)
			ans[i] = arr[i].r > arr[i - 1].r ? i : ans[i - 1];
		for (int i = 0; i < k; i++) {
			int x = convert[in.nextInt() - 1];
			int y = convert[in.nextInt() - 1];
			if (x < y && arr[x].r < arr[y].r)
				ans[y]--;
			else if (y < x && arr[y].r < arr[x].r)
				ans[x]--;
		}
		
		for (int i = 0; i < n; i++)
			System.out.print(ans[convert[i]] + (i == n - 1 ? "\n" : " "));
	}
	
	public static class Programmer implements Comparable<Programmer>
	{
		int i;
		int r;
		
		public Programmer(int a, int b)
		{
			i = a;
			r = b;
		}
		
		public int compareTo(Programmer o)
		{
			return this.r - o.r;
		}
	}
	
	/**
	 * Source: Matt Fontaine
	 */
	public static class FastScanner
	{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public FastScanner(InputStream stream)
		{
			this.stream = stream;
		}

		int read()
		{
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars)
			{
				curChar = 0;
				try
				{
					numChars = stream.read(buf);
				} catch (IOException e)
				{
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		boolean isSpaceChar(int c)
		{
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		boolean isEndline(int c)
		{
			return c == '\n' || c == '\r' || c == -1;
		}

		public int nextInt()
		{
			return Integer.parseInt(next());
		}

		public long nextLong()
		{
			return Long.parseLong(next());
		}

		public double nextDouble()
		{
			return Double.parseDouble(next());
		}

		public String next()
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine()
		{
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
	}
}