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 sys,math
from collections import deque,defaultdict
import operator as op
from functools import reduce
from itertools import permutations

# sys.setrecursionlimit(10**6) 
#OneDrive\Documents\codeforces

I=sys.stdin.readline

alpha="abcdefghijklmnopqrstuvwxyz"

"""
x_move=[-1,0,1,0,-1,1,1,-1]
y_move=[0,1,0,-1,1,1,-1,-1]
"""
def ii():
	return int(I().strip())
def li():
	return list(map(int,I().strip().split()))
def mi():
	return map(int,I().strip().split())


def ncr(n, r):
    r = min(r, n-r)
    numer = reduce(op.mul, range(n, n-r, -1), 1)
    denom = reduce(op.mul, range(1, r+1), 1)
    return numer // denom 
 

def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

def isPrime(n):
	if n<=1:
		return False
	elif n<=2:
		return True
	else:
		
		for i in range(2,int(n**.5)+1):
			if n%i==0:
				return False
		return True


#print("Case #"+str(_+1)+":",abs(cnt-k))




def main():
	
	n,k=mi()
	arr=li()
	d=defaultdict(list)

	ans=[0]*n
	for i in range(k):
		u,v=mi()
		if arr[u-1]>arr[v-1]:
			ans[u-1]-=1
		if arr[v-1]>arr[u-1]:
			ans[v-1]-=1 
	tmp=[i for i in arr]
	tmp.sort()

	# ans=[]
	# print(tmp)
	for i in range(n):
		low=0
		high=n-1

		while low<high:
			mid=low+(high-low)//2

			if tmp[mid]>=arr[i]:
				high=mid
			else:
				low=mid+1
		ans[i]+=high 

	print(*ans)




		
	

					




		

		


			





		



	

	

	

	




			

	





	








							





		

	

	

		
				
					

		




		










	






		

	



	








		
			





		






	

if __name__ == '__main__':
	main()