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.
# Python 3 implementation of the approach 

# Function to return the count of elements 
# from the array which are greater than k 
def countGreater(arr, n, k): 
	l = 0
	r = n - 1

	# Stores the index of the left most element 
	# from the array which is greater than k 
	leftGreater = n 

	# Finds number of elements greater than k 
	while (l <= r): 
		m = int(l + (r - l) / 2) 

		# If mid element is greater than 
		# k update leftGreater and r 
		if (arr[m] > k): 
			leftGreater = m 
			r = m - 1

		# If mid element is less than 
		# or equal to k update l 
		else: 
			l = m + 1

	# Return the count of elements 
	# greater than k 
	return (n - leftGreater) 
	
n,k=map(int,input().split())
r=input().split()
ls=[]
for i in range(n):
    r[i]=int(r[i])
    ls.append(-r[i])
ls.sort()
ans=[countGreater(ls,n,-r[i]) for i in range(n)]
for i in range(k):
    u,v=map(int,input().split())
    if(r[u-1]>r[v-1]):
        ans[u-1]-=1
    elif(r[v-1]>r[u-1]):
        ans[v-1]-=1
print(*ans)        