#include <bits/stdc++.h>
using namespace std;
long long int calc(char s[], long long int begin, long long int limit,
                   long long int end) {
  long long int i, j, minimum = end + 1;
  for (i = begin; i < limit; i++) {
    for (j = 1; j <= (limit - i) / 2; j++) {
      if (s[i] == s[i + j] && s[i] == s[i + 2 * j]) {
        if (i + 2 * j <= minimum) {
          minimum = i + 2 * j;
        }
      }
    }
  }
  return (end - minimum + 1);
}
int main() {
  char s[300001];
  long long int n, i, count = 0;
  scanf("%s", s + 1);
  i = 1;
  while (s[i] == '0' || s[i] == '1') {
    i++;
  }
  n = i - 1;
  for (i = 1; i <= n; i++) {
    if (i + 8 <= n)
      count += calc(s, i, i + 8, n);
    else
      count += calc(s, i, n, n);
  }
  printf("%lld\n", count);
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int calc(char s[], long long int begin, long long int limit,
                   long long int end) {
  long long int i, j, minimum = end + 1;
  for (i = begin; i < limit; i++) {
    for (j = 1; j <= (limit - i) / 2; j++) {
      if (s[i] == s[i + j] && s[i] == s[i + 2 * j]) {
        if (i + 2 * j <= minimum) {
          minimum = i + 2 * j;
        }
      }
    }
  }
  return (end - minimum + 1);
}
int main() {
  char s[300001];
  long long int n, i, count = 0;
  scanf("%s", s + 1);
  i = 1;
  while (s[i] == '0' || s[i] == '1') {
    i++;
  }
  n = i - 1;
  for (i = 1; i <= n; i++) {
    if (i + 8 <= n)
      count += calc(s, i, i + 8, n);
    else
      count += calc(s, i, n, n);
  }
  printf("%lld\n", count);
  return 0;
}

```