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.*;
import java.io.*;
public class Main
{
    public static void main(String [] args)throws IOException
    {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        String s=sc.next();
        int store[]=new int[s.length()];
        for(int i=0;i<s.length();i++)
            store[i]=1000000000;
        ArrayList<Integer> al[]=new ArrayList[2];
        for(int i=0;i<2;i++)
            al[i]=new ArrayList();
        for(int i=0;i<s.length();i++)
        {
            if(s.charAt(i)=='0')
                al[0].add(i);
            else
                al[1].add(i);
        }
        for(int k=0;k<2;k++)
            for(int i=0;i<al[k].size();i++)
            {
                for(int j=i+1;j<al[k].size();j++)
                {
                    try
                    {
                        if (s.charAt(2 * al[k].get(j) - al[k].get(i)) == (k==0?'0':'1'))
                        {
                            store[al[k].get(i)] = 2 * al[k].get(j) - al[k].get(i);
                            break;
                        }
                    }
                    catch(Exception e)
                    {
                        ;
                    }
                }
            }
        int biggest=s.length();
        long ans=0;
        for(int i=s.length()-1;i>=0;i--)
        {
            if(store[i]>s.length())
                continue;
            if(biggest>store[i])
            {
                ans += (i + 1) * (long) (biggest - store[i]);
                biggest = store[i];
            }
        }
        out.println(ans);
        out.flush();
    }
}