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 bisect

# def binary_search_lower_bound(target, num):
#     l = 0
#     r = len(num) - 1
#     while l <= r:
#         mid = (l + r) // 2
#         if target < num[mid]:
#             r = mid - 1
#         elif target > num[mid]:
#             l = mid + 1
#         else:
#             if mid > 0 and num[mid - 1] == target:
#                 r = mid - 1
#             else:
#                 l = mid
#                 break
#     return l

if __name__ == '__main__':
    firstLineNum = [int(x) for x in input().split(' ')]
    n = firstLineNum[0]
    k = firstLineNum[1]

    skill = [int(x) for x in input().split(' ')]
    ans = [0] * len(skill)

    sortedSkill = list(skill)
    sortedSkill.sort()

    for i in range(k):
        numList = [int(x) for x in input().split(' ')]
        x = numList[0]
        y = numList[1]
        if skill[x - 1] > skill[y - 1]:
            ans[x - 1] -= 1
        elif skill[x - 1] < skill[y - 1]:
            ans[y - 1] -= 1

    for index, p in enumerate(skill):
        # pos = binary_search_lower_bound(p, sortedSkill)
        pos = bisect.bisect_left(sortedSkill, p)
        print(pos + ans[index], end=' ')