#include <bits/stdc++.h>
using namespace std;
string s;
int fin(int i1) {
  int mi = 999999999;
  for (int d = 1; d <= 9; d++) {
    for (int i = i1; i < i1 + 10; i++) {
      if (i + 2 * d < s.length()) {
        if (s[i] == s[i + d] && s[i] == s[i + 2 * d]) {
          mi = min(mi, i + 2 * d);
        }
      }
    }
  }
  return mi;
}
int main() {
  cin >> s;
  long long n = s.length();
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int p = fin(i);
    if (n > p) {
      ans = ans + n - p;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 fin(int i1) {
  int mi = 999999999;
  for (int d = 1; d <= 9; d++) {
    for (int i = i1; i < i1 + 10; i++) {
      if (i + 2 * d < s.length()) {
        if (s[i] == s[i + d] && s[i] == s[i + 2 * d]) {
          mi = min(mi, i + 2 * d);
        }
      }
    }
  }
  return mi;
}
int main() {
  cin >> s;
  long long n = s.length();
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int p = fin(i);
    if (n > p) {
      ans = ans + n - p;
    }
  }
  cout << ans;
  return 0;
}

```