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 bisect
# import heapq
from math import ceil,floor


def check(k,n):
    fr = 0
    temp=  n
    while ( n > 0):
        if n < k :
            fr = fr+n
            n = 0
        else:
            fr = fr + k
            n = n-k
            n -= (n//10)
    return fr*2>= temp

RI = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
mod = 10**9+7
# for _ in range(int(ri())):

n,k = RI()
a = RI()
lis = [[] for i in range(n)]
for i in range(k):
    l,r = RI()
    
    if a[l-1] < a[r-1]:
        lis[r-1].append(l-1)
    elif a[l-1] > a[r-1]:
        lis[l-1].append(r-1)

sor = [[a[i],i] for i in range(len(a))]
sor.sort(key = lambda x: x[0])

cnt = {}
seen = set()
ans = [0]*n
# print(lis,sor)
for i in range(len(sor)):
    
    if sor[i][0] not in cnt:
        cnt[sor[i][0]] = 1
    else:
        cnt[sor[i][0]]+=1
    
    count = i-(cnt[sor[i][0]] -1)
    for j in lis[sor[i][1]]:
        if j in seen:
            count-=1
    # print(count)
    ans[sor[i][1]] = count
    seen.add(sor[i][1])
for i in ans:
    print(i,end=" ")
print()