CF978F · Mentors

• 难度：1500
• 标签：binary search、data structures、implementation
• 链接：https://codeforces.com/problemset/problem/978/F
• 时间限制：3 seconds　内存限制：256 megabytes
• 出现位置：Day27-阶段模拟6-S T1T2稳分卷

英文原题面

Statement

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 $b$ $(r_a \gt r_b)$ 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.

样例

样例 1

输入：

4 2
10 4 10 15
1 2
4 3

输出：

0 0 1 2 

样例 2

输入：

10 4
5 4 1 5 4 3 7 1 2 5
4 6
2 1
10 8
3 5

输出：

5 4 0 5 3 3 9 0 2 5 

样例解释（英文原文）

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.