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 Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val s = readLine()!!
    var cnt = 0L
    for (i in 0 until n-2) {
        if (s[i] == 'R') {
            for (j in i+1 until n-1) {
                if (s[j] == 'R') {
                    continue
                } else if (s[j] == 'G') {
                    for (k in j+1 until n) {
                        if (s[k] == 'B' && j-i != k-j) cnt++
                    }
                } else {
                    for (k in j+1 until n) {
                        if (s[k] == 'G' && j-i != k-j) cnt++
                    }                    
                }
            }
        } else if (s[i] == 'G') {
            for (j in i+1 until n-1) {
                if (s[j] == 'G') {
                    continue
                } else if (s[j] == 'R') {
                    for (k in j+1 until n) {
                        if (s[k] == 'B' && j-i != k-j) cnt++
                    }
                } else {
                    for (k in j+1 until n) {
                        if (s[k] == 'R' && j-i != k-j) cnt++
                    }
                }
            }
        } else {
            for (j in i+1 until n-1) {
                if (s[j] == 'B') {
                    continue
                } else if (s[j] == 'R') {
                    for (k in j+1 until n) {
                        if (s[k] == 'G' && j-i != k-j) cnt++
                    }
                } else {
                    for (k in j+1 until n) {
                        if (s[k] == 'R' && j-i != k-j) cnt++
                    }
                }
            }
        }
    }
    println(cnt)
}