#include <bits/stdc++.h>
using namespace std;
ifstream f("1.in");
long long len;
string s;
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) len += i;
  for (int i = 0; i < s.size(); i++) {
    for (int j = i - 1; j >= 0; j--) {
      int gasit = 0;
      for (int x = 0; x <= 10 && gasit == 0; x++)
        for (int k = 1; k <= 10; k++) {
          if (j + x + 2 * k > i) break;
          if (s[j + x] == s[j + x + k] && s[j + x + k] == s[j + x + 2 * k])
            gasit++;
        }
      if (gasit) break;
      len--;
    }
  }
  cout << len;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
ifstream f("1.in");
long long len;
string s;
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) len += i;
  for (int i = 0; i < s.size(); i++) {
    for (int j = i - 1; j >= 0; j--) {
      int gasit = 0;
      for (int x = 0; x <= 10 && gasit == 0; x++)
        for (int k = 1; k <= 10; k++) {
          if (j + x + 2 * k > i) break;
          if (s[j + x] == s[j + x + k] && s[j + x + k] == s[j + x + 2 * k])
            gasit++;
        }
      if (gasit) break;
      len--;
    }
  }
  cout << len;
  return 0;
}

```