#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 5;
int minhold[MAXN];
long long ans;
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    minhold[i] = 1e9;
    for (int k = 1; i + 2 * k < s.length(); k++) {
      if (s[i] == s[i + k] && s[i + k] == s[i + 2 * k]) {
        minhold[i] = i + 2 * k;
        break;
      }
    }
  }
  int tempmin = 1e9;
  for (int i = s.length() - 1; i >= 0; i--) {
    tempmin = min(tempmin, minhold[i]);
    if (tempmin < s.length()) {
      ans += (s.length() - tempmin);
    }
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
const int MAXN = 3e5 + 5;
int minhold[MAXN];
long long ans;
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < s.length(); i++) {
    minhold[i] = 1e9;
    for (int k = 1; i + 2 * k < s.length(); k++) {
      if (s[i] == s[i + k] && s[i + k] == s[i + 2 * k]) {
        minhold[i] = i + 2 * k;
        break;
      }
    }
  }
  int tempmin = 1e9;
  for (int i = s.length() - 1; i >= 0; i--) {
    tempmin = min(tempmin, minhold[i]);
    if (tempmin < s.length()) {
      ans += (s.length() - tempmin);
    }
  }
  cout << ans << endl;
}

```