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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<algorithm>//sort,二分探索,など
#include<bitset>//固定長bit集合
#include<cmath>//pow,logなど
#include<complex>//複素数
#include<deque>//両端アクセスのキュー
#include<functional>//sortのgreater
#include<iomanip>//setprecision(浮動小数点の出力の誤差)
#include<iostream>//入出力
#include<map>//map(辞書)
#include<numeric>//iota(整数列の生成),gcdとlcm(c++17)
#include<queue>//キュー
#include<set>//集合
#include<stack>//スタック
#include<string>//文字列
#include<unordered_map>//イテレータあるけど順序保持しないmap
#include<unordered_set>//イテレータあるけど順序保持しないset
#include<utility>//pair
#include<vector>//可変長配列
using namespace std;

typedef long long ll;

//マクロ
#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define REPD(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define FOR(i,a,b) for(ll i=(a);i<=(b);i++)
#define FORD(i,a,b) for(ll i=(a);i>=(b);i--)
#define ALL(x) (x).begin(),(x).end() //sortなどの引数を省略したい
#define SIZE(x) ((ll)(x).size()) //sizeをsize_tからllに直しておく
#define MAX(x) *max_element(ALL(x))
#define INF 1000000000000 //10^12
#define MOD 10000007 //10^9+7
#define MAXR 100000 //10^5:最大のrange(素数列挙などで使用)


int main(void){
    int n;
    cin >> n;
    string s;
    cin >> s;
    vector<int> a, b, c;
    REP(i, n) {
        if(s[i] == 'R') a.push_back(i+1);
        if(s[i] == 'G') b.push_back(i+1);
        if(s[i] == 'B') c.push_back(i+1);
    }
    ll sum = a.size()*b.size()*c.size();
    
    //とびとびのものを数えて引く
    for(int i=1; i<n; i++) {
        for(int j=0; j+i+i<n; j++) {
            if(s[j] == s[j+i] || s[j+i] == s[j+i+i] || s[j] == s[j+i+i]) continue;
            sum--;
        }
    }

    cout << sum << endl;
    return 0;
}