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.
"""Problem F - Mentors.

http://codeforces.com/contest/978/problem/f

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, \frac{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.

"""
import logging
from collections import defaultdict


fmt = '%(levelname)s - %(name)s (line:%(lineno)s) - %(message)s'
formatter = logging.Formatter(fmt)

ch = logging.StreamHandler()
ch.setLevel(logging.ERROR)
ch.setFormatter(formatter)

logger = logging.getLogger('mentors')
logger.setLevel(logging.ERROR)
logger.addHandler(ch)


def solve(skills, quarrels):
    g1 = defaultdict(list)
    for i, skill in enumerate(skills):
        g1[skill].append(i)
    
    g2 = defaultdict(list)
    for a, b in quarrels:
        if skills[a] == skills[b]:
            continue
        if skills[a] < skills[b]:
            a, b = b, a
        g2[a].append(b)

    n = len(skills)
    ans = [0] * len(skills)
    for skill in sorted(set(skills), reverse=True):
        for i in g1[skill]:
            ans[i] = max(0, n - len(g2[i]) - len(g1[skill]))
        n -= len(g1[skill])
        del g1[skill]

    return ans


def main():
    _, k = map(int, input().strip().split())
    skills = [int(x) for x in input().strip().split()]

    quarrels = []
    for _ in range(k):
        a, b = map(int, input().strip().split())
        a, b = a - 1, b - 1
        quarrels.append((a, b))
    
    result = solve(skills, quarrels)
    print(' '.join(map(str, result)))
    

if __name__ == '__main__':
    main()
