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 collections import Counter
def binary(x,ar):
    low=0
    high=len(ar)-1
    while low<=high:
        mid=(low+high)//2
        if ar[mid]==x:
            break
        elif ar[mid]<x:
            low=mid+1
        else:
            high=mid-1
    return mid


n,q=map(int,input().split())
l=list(map(int,input().split()))
frq=dict(Counter(l))
new=sorted(set(l))
look=[frq[new[0]]]
for i in range(1,len(new)):
    look.append(look[i-1]+frq[new[i]])
d=dict()
for i in range(q):
    li,r=map(int,input().split())
    li-=1
    r-=1
    if li in d:
        d[li].append(r)
    else:
        d[li]=[r]
    
    if r in d:
        d[r].append(li)
    else:
        d[r]=[li]
enum=[]
for i in range(n):
    x=binary(l[i],new)
    if x==0:
        sum1=0
    else:
        sum1=look[x-1]
        if i in d:
            
            for j in range(len(d[i])):
                if l[d[i][j]]<l[i]:
                    sum1-=1
    enum.append(sum1)
print(*enum)
    
    
    
            

