Problem:
Score : 400 points

Problem Statement

We have a string S of length N consisting of R, G, and B.

Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:

S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.

j - i \neq k - j.

Constraints

1 \leq N \leq 4000

S is a string of length N consisting of R, G, and B.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of triplets in question.

Sample Input 1

4
RRGB

Sample Output 1

1

Only the triplet (1,~3,~4) satisfies both conditions. The triplet (2,~3,~4) satisfies the first condition but not the second, so it does not count.

Sample Input 2

39
RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB

Sample Output 2

1800

Solution in Perl (5.26.1) (judge verdict: Time Limit Exceeded)

use v5.26; # strict say state biwise
use warnings;
use List::Util qw(reduce any all none notall first max maxstr min minstr product sum sum0 pairs unpairs pairkeys pairvalues pairfirst pairgrep pairmap shuffle uniq uniqnum uniqstr);

chomp (my $n = <STDIN>);
chomp (my $s = <STDIN>);
my @s = (0, split('', $s));
my (@r, @g, @b);
for my $i (1..$n) {
  if ($s[$i] eq 'R') {
    unshift @r, $i;
  } elsif($s[$i] eq 'G') {
    unshift @g, $i;
  } else {
    unshift @b, $i;
  }
}
my $ans = 0;
for my $i (@r) {
  for my $j (@g) {
    last if $i > $j;
    for my $k (@b) {
      last if $j > $k;
      $ans++ if $j - $i != $k - $j;
    }
  }
}
for my $i (@r) {
  for my $j (@b) {
    last if $i > $j;
    for my $k (@g) {
      last if $j > $k;
      $ans++ if $j - $i != $k - $j;
    }
  }
}
for my $i (@g) {
  for my $j (@r) {
    last if $i > $j;
    for my $k (@b) {
      last if $j > $k;
      $ans++ if $j - $i != $k - $j;
    }
  }
}
for my $i (@g) {
  for my $j (@b) {
    last if $i > $j;
    for my $k (@r) {
      last if $j > $k;
      $ans++ if $j - $i != $k - $j;
    }
  }
}
for my $i (@b) {
  for my $j (@r) {
    last if $i > $j;
    for my $k (@g) {
      last if $j > $k;
      $ans++ if $j - $i != $k - $j;
    }
  }
}
for my $i (@b) {
  for my $j (@g) {
    last if $i > $j;
    for my $k (@r) {
      last if $j > $k;
      $ans++ if $j - $i != $k - $j;
    }
  }
}
say $ans;