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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;

namespace AtCoder.D
{
    internal static class CR
    {
        public static string GetText() { return Console.ReadLine(); }
        public static List<string> GetTexts() { return Console.ReadLine().Split(' ').ToList(); }
        public static long GetLong() { return long.Parse(Console.ReadLine()); }
        public static List<long> GetLongs() { return Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToList(); }
    }

    public class Program
    {
        public static long N { get; private set; }
        public static string S { get; private set; }

        public static void Main() { var r = GetResult(); Debug.WriteLine(r); Console.Write(r); }

        private static object GetResult()
        {
            N = CR.GetLong();
            S = CR.GetText();

            var rCount = 0;
            var gCount = 0;
            var bCount = 0;
            var sameDisCount = 0;
            for (int i = 0; i < N; i++)
            {
                switch (S[i])
                {
                    case 'R': rCount++; break;
                    case 'G': gCount++; break;
                    case 'B': bCount++; break;
                    default:
                        break;
                }

                for (int j = i + 1; j < N; j++)
                {
                    if (S[i] == S[j]) continue;
                    var k = 2 * j - i;
                    if (k < N && k > j)
                    {
                        if (S[i] != S[k] && S[j] != S[k])
                        {
                            sameDisCount++;
                        }
                    }
                }
            }

            return rCount * gCount * bCount - sameDisCount;
        }
    }
}