#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    string s;
    cin >> s;
    long long int n = s.size();
    long long int w = 0, x = 0, p;
    long long int ans = 0;
    for (int i = 0; i < n - 2; i++) {
      w = 0;
      x = min(n, i + 13LL);
      long long int c = 0;
      for (int j = x; j >= i + 2; j--) {
        for (int k = j - 2; k >= i; k--) {
          if (s[j] == s[k] && (j + k) % 2 == 0 && s[(j + k) / 2] == s[j]) {
            c = max(c, n - j);
          }
        }
        if (w == 1) break;
      }
      ans += c;
    }
    cout << ans;
  }
  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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    string s;
    cin >> s;
    long long int n = s.size();
    long long int w = 0, x = 0, p;
    long long int ans = 0;
    for (int i = 0; i < n - 2; i++) {
      w = 0;
      x = min(n, i + 13LL);
      long long int c = 0;
      for (int j = x; j >= i + 2; j--) {
        for (int k = j - 2; k >= i; k--) {
          if (s[j] == s[k] && (j + k) % 2 == 0 && s[(j + k) / 2] == s[j]) {
            c = max(c, n - j);
          }
        }
        if (w == 1) break;
      }
      ans += c;
    }
    cout << ans;
  }
  return 0;
}

```