#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 4;
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  cin >> s;
  long long n = s.size();
  long long sub = 0;
  for (int k = 1; k <= 8; k++) {
    for (int i = 0; i + k <= n; i++) {
      bool flag = false;
      for (int j = 1; 2 * j < k; j++) {
        for (int st = i; st + 2 * j < i + k; st++) {
          if (s[st] == s[st + j] && s[st] == s[st + 2 * j]) {
            flag = true;
          }
        }
      }
      if (!flag) sub++;
    }
  }
  cout << n * (n + 1) / 2 - sub << '\n';
}

### 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 MAX = 2e5 + 4;
string s;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  cin >> s;
  long long n = s.size();
  long long sub = 0;
  for (int k = 1; k <= 8; k++) {
    for (int i = 0; i + k <= n; i++) {
      bool flag = false;
      for (int j = 1; 2 * j < k; j++) {
        for (int st = i; st + 2 * j < i + k; st++) {
          if (s[st] == s[st + j] && s[st] == s[st + 2 * j]) {
            flag = true;
          }
        }
      }
      if (!flag) sub++;
    }
  }
  cout << n * (n + 1) / 2 - sub << '\n';
}

```