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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;

public class D {
	public static void main(String[] args) throws IOException {
		BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
		HashSet<String> stringSet = new HashSet<String>();
		
		for (int l = 1; l <= 8; l++) {
			for (int i = 0; i < 1 << l; i++) {
				boolean found = false;
				for (int a = 0; a < l; a++) {
					for (int k = 1; a + 2 * k < l; k++) {
						int b = a + k;
						int c = a + 2 * k;
						
						int aBit = (i & (1 << a)) >> a;
						int bBit = (i & (1 << b)) >> b;
						int cBit = (i & (1 << c)) >> c;
						
						if (aBit == bBit && bBit == cBit) {
							found = true;
						}
					}
				}
				
				if (!found) {
					StringBuilder sb = new StringBuilder();
					for (int j = 0; j < l; j++) {
						int bit = (i & (1 << j)) >> j;
						sb.append(bit);
					}
					stringSet.add(sb.reverse().toString());
				}
			}
		}
		
		String binaryString = input.readLine();
		long totalWays = ((long) binaryString.length()) * (binaryString.length() + 1) / 2;
		//System.out.println("Ways: " + totalWays);
		for (int i = 0; i <= binaryString.length(); i++) {
			for (int j = 2; i - j + 1 >= 0 && j <= 9; j++) {
				String substring = binaryString.substring(i - j + 1, i);
				//System.out.println(substring);
				if (stringSet.contains(substring)) {
					totalWays--;
				}
			}
			//System.out.println("Ways: " + totalWays);
		}
		
		System.out.println(totalWays);
	}
}