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
import math
import collections
import sys

from functools import reduce
import decimal

sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP

graphDict = collections.defaultdict
queue = collections.deque


class Graphs:
    def __init__(self):
        self.graph = graphDict(list)

    def add_edge(self, u, v):
        self.graph[u].append(v)

    def dfs_utility(self, nodes, visited_nodes):
        visited_nodes.add(nodes)
        for neighbour in self.graph[nodes]:
            if neighbour not in visited_nodes:
                self.dfs_utility(neighbour, visited_nodes)
            else:
                return neighbour

    def dfs(self, node):
        Visited = set()
        self.dfs_utility(node, Visited)

    def bfs(self, node):
        visited = set()
        if node not in visited:
            queue.append(node)
            visited.add(node)
        while queue:
            parent = queue.popleft()
            print(parent)
            for item in self.graph[parent]:
                if item not in visited:
                    queue.append(item)
                    visited.add(item)


def rounding(n):
    return int(decimal.Decimal(f'{n}').to_integral_value())


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


def inp():
    return sys.stdin.readline().strip()


def map_inp(v_type):
    return map(v_type, inp().split())


def list_inp(v_type):
    return list(map_inp(v_type))


######################################## Solution ####################################


n, k = map_inp(int)
arr = list_inp(int)
arr_copy = sorted(arr)
result = [bisect.bisect_left(arr_copy, item) for item in arr]
for i in range(k):
    a, b = map_inp(int)
    if arr[a - 1] < arr[b - 1]:
        result[b - 1] -= 1
    elif arr[a - 1] > arr[b - 1]:
        result[a - 1] -= 1
print(*result)
