#include <bits/stdc++.h>
using namespace std;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin >> s;
  long long ans = 0;
  int n = s.length();
  for (int i = 0; i < n; ++i) {
    bool is = 0;
    for (int j = i + 2; j < n; ++j) {
      for (int k = 1; k <= 4; ++k) {
        if (j - 2 * k >= i) {
          if (s[j] == s[j - k] && s[j] == s[j - 2 * k]) {
            ans += n - j;
            is = 1;
            break;
          }
        }
      }
      if (is) break;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin >> s;
  long long ans = 0;
  int n = s.length();
  for (int i = 0; i < n; ++i) {
    bool is = 0;
    for (int j = i + 2; j < n; ++j) {
      for (int k = 1; k <= 4; ++k) {
        if (j - 2 * k >= i) {
          if (s[j] == s[j - k] && s[j] == s[j - 2 * k]) {
            ans += n - j;
            is = 1;
            break;
          }
        }
      }
      if (is) break;
    }
  }
  cout << ans << endl;
  return 0;
}

```