Problem: Toad Rash has a binary string s. A binary string consists only of zeros and ones.

Let n be the length of s.

Rash needs to find the number of such pairs of integers l, r that 1 ≤ l ≤ r ≤ n and there is at least one pair of integers x, k such that 1 ≤ x, k ≤ n, l ≤ x < x + 2k ≤ r, and s_x = s_{x+k} = s_{x+2k}.

Find this number of pairs for Rash.

Input

The first line contains the string s (1 ≤ |s| ≤ 300 000), consisting of zeros and ones.

Output

Output one integer: the number of such pairs of integers l, r that 1 ≤ l ≤ r ≤ n and there is at least one pair of integers x, k such that 1 ≤ x, k ≤ n, l ≤ x < x + 2k ≤ r, and s_x = s_{x+k} = s_{x+2k}.

Examples

Input


010101


Output


3


Input


11001100


Output


0

Note

In the first example, there are three l, r pairs we need to count: 1, 6; 2, 6; and 1, 5.

In the second example, there are no values x, k for the initial string, so the answer is 0.

Solution (in JAVA):

Solution:


import java.util.Scanner;
//可以知道01字符串在长度很短的时候就一定存在这样的区间，所以就算暴力枚举左端点，满足条件的右端点也一定会在左端点不远处，
//所以可以直接暴力枚举左端点i，一点点检查恰好成立的右端点j，则有ans+=len-j
public class Main {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		String s=sc.nextLine();
		char ctr[]=s.toCharArray();
		int len=ctr.length;
		long ans=0;
		for(int i=0;i<len;i++){//枚举左端点
			boolean flag=false;
			for(int j=i;j<len;j++){//寻找满足条件的右端点
				for(int k=1;j-2*k>=i;k++){
					if(ctr[j]==ctr[j-k]&&ctr[j-k]==ctr[j-2*k]){
						ans=ans+len-j;
						flag=true;
						break;
					}
				}
				if(flag)
					break;
			}
		}
		
		System.out.println(ans);
		sc.close();
	}
}