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.
from sys import stdin, stdout 
from bisect import bisect_left, bisect_right
from collections import defaultdict
import math
cin = stdin.readline
def cout(x):
	stdout.write(str(x)+'\n')

def nexint():
    return int(stdin.readline())
def readline():
    return map(int,stdin.readline().split())
def readlist():
    return list(map(int,stdin.readline().split()))

def sorted_indexes(arr):
    return sorted(range(len(arr)),key=arr.__getitem__)
def printr(arr):
    [stdout.write(str(x)+' ')   for x in arr]
    cout('')

def find_lt(a, x):#'Find rightmost value less than x'
    i = bisect_left(a, x)
    if i:
        return a[i-1]
    raise ValueError

def find_gt(a, x):#'Find leftmost value greater than x'
    i = bisect_right(a, x)
    if i != len(a):
        return a[i]
    raise ValueError

# ---------------------Template ends-------------------------------------
n,k = readline()
arr = readlist()

xx = defaultdict(list)
for _ in range(k):
    x, y = readline()
    x-=1
    y-=1
    xx[x].append(y)
    xx[y].append(x)
clone = arr[:]
ix = sorted_indexes(arr)
arr.sort()
ans = [0]*n
for i in range(n):
    z = bisect_left(arr, clone[i])
    if(z==0):
        ans[i] = 0
        continue
    
    for index in xx[i]:
        if clone[index] < clone[i]:
            z-=1    
    ans[i] = z

printr(ans)

