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 stdout, stdin, setrecursionlimit
from io import BytesIO, IOBase
from collections import *
from itertools import *
from random import *
from bisect import *
from string import *
from queue import *
from heapq import *
from math import *
from re import *
from os import *

####################################---fast-input-output----#########################################


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = read(self._fd, max(fstat(self._fd).st_size, 8192))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = read(self._fd, max(fstat(self._fd).st_size, 8192))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
def fast(): return stdin.readline().strip()
def zzz(): return [int(i) for i in fast().split()]


z, zz = fast, lambda: list(map(int, z().split()))
szz, graph, mod, szzz = lambda: sorted(
    zz()), {}, 10**9 + 7, lambda: sorted(zzz())


def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
def output(answer, end='\n'): stdout.write(str(answer) + end)


dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]


#################################################---Some Rule For Me To Follow---#################################
"""
    --instants of Reading problem continuously try to understand them.

    --If you Know some-one , Then you probably don't know him !

    --Try & again try


"""
##################################################---START-CODING---###############################################

n,m = zzz()
arr = zzz()
newArr = [[i,j] for j,i in enumerate(arr)]

newArr = sorted(newArr)



for _ in range(m):
    x,y=zzz()
    x-=1
    y-=1
    try:
        graph[x].append(y)
    except:
        graph[x]=[y]
    try:
        graph[y].append(x)
    except:
        graph[y]=[x]


new={}
cnt=0
# print(newArr)
for i in range(n):
    if i!=0 and newArr[i][0]==newArr[i-1][0]:
        cnt+=1
    else:
        cnt=0
    new[newArr[i][1]]=i-cnt

# print(new)
ans= [0]*n

for i in range(n):
    idx=newArr[i][1]
    num = newArr[i][0]
    res = new[idx]
    # print(res,idx)
    try:
        for j in graph[idx]:
            if num>arr[j]:
                res-=1
    except:
        pass
    ans[idx]=res
print(*ans)