# Mentors

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 $k$ $(2 \le n \le 2 \cdot 10^5$, $0 \le k \le \min(2 \cdot 10^5, \frac{n \cdot (n – 1)}{2}))$ — total number of programmers and number of pairs of programmers which are in a quarrel.

The second line contains a sequence of integers $r_1, r_2, \dots, r_n$ $(1 \le r_i \le 10^{9})$, where $r_i$ equals to the skill of the $i$-th programmer.

Each of the following $k$ lines contains two distinct integers $x$, $y$ $(1 \le x, y \le n$, $x \ne y)$ — 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.Examplesinput

4 210 4 10 151 24 3

output

0 0 1 2

input

10 45 4 1 5 4 3 7 1 2 54 62 110 83 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.

Solution:

private fun readLn() = readLine()!! // string line
private fun readInt() = readLn().toInt() // single int
private fun readLong() = readLn().toLong() // single long
private fun readDouble() = readLn().toDouble() // single double
private fun readStrings() = readLn().split(" ") // list of strings
private fun readInts() = readStrings().map { it.toInt() } // list of ints
private fun readLongs() = readStrings().map { it.toLong() } // list of longs
private fun readDoubles() = readStrings().map { it.toDouble() } // list of doubles

fun main(args: Array<String>) {
var (n, k) = readInts()
var r = readInts()
var order = IntArray(n) {it}.sortedBy {r[it]}
var ans = IntArray(n)
var beg = 0
while (beg < n) {
var end = beg
while (end + 1 < n && r[order[end + 1]] == r[order[end]]) {
end += 1
}
for (i in beg..end) {
ans[order[i]] = beg
}
beg = end + 1
}
for (i in 1..k) {
var (x, y) = readInts()
--x
--y
if (r[x] > r[y]) {
--ans[x]
}
if (r[x] < r[y]) {
--ans[y]
}
}
println(ans.joinToString(" "))
}