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

### Prompt

Generate 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(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = (int)s.size();
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int j = n;
    for (int x = i; x < n; x++) {
      for (int k = 1; k <= n && x + 2 * k < n; k++) {
        if (s[x] == s[x + k] && s[x + k] == s[x + 2 * k]) {
          j = min(j, x + 2 * k);
          break;
        }
      }
      if (x > j) {
        break;
      }
    }
    ans += n - j;
  }
  cout << ans;
}

```