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
import heapq
import bisect
import math
import random

INF = 10**9+7
OFFLINE = 0
N = 101010
sys.setrecursionlimit(INF)

def fi():
	return int(sys.stdin.readline())

def fi2():
	return map(int, sys.stdin.readline().split())

def fi3():
	return sys.stdin.readline().rstrip()

def fo(*args):
	for s in args:
		sys.stdout.write(str(s)+" ")
	sys.stdout.write("\n")

##
if OFFLINE:
	sys.stdin = open("fin.txt", "r")
	sys.stdout = open("fout.txt", "w")
##



##main

n, k = fi2()

a = fi2()

res = {}
for i in range(n):
	res[i] = 0

for i in range(k):
	u, v = fi2()

	if a[u-1] < a[v-1]:
		res[v-1] -= 1

	if a[u-1] > a[v-1]:
		res[u-1] -= 1


b = sorted(a)

for i in range(n):
	index = bisect.bisect_left(b, a[i])

	res[i] += index

ans = []

for i in range(n):
	ans.append(str(res[i]))

ans = ' '.join(ans)

fo(ans)