#include <bits/stdc++.h>
using namespace std;
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  n = (long long)s.length();
  long long cnt = 0;
  for (long long l = 0; l < n - 2; l++) {
    bool nemu = 0;
    for (long long r = l + 2; r < n; r++) {
      for (long long k = 1; k + k + l <= r; k++) {
        for (long long mul = l; mul + k + k <= r; mul++) {
          if (s[mul] == s[mul + k] && s[mul] == s[mul + k + k]) {
            nemu = 1;
            break;
          }
        }
        if (nemu) break;
      }
      if (nemu) {
        cnt += n - r;
        break;
      }
    }
  }
  cout << cnt << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  n = (long long)s.length();
  long long cnt = 0;
  for (long long l = 0; l < n - 2; l++) {
    bool nemu = 0;
    for (long long r = l + 2; r < n; r++) {
      for (long long k = 1; k + k + l <= r; k++) {
        for (long long mul = l; mul + k + k <= r; mul++) {
          if (s[mul] == s[mul + k] && s[mul] == s[mul + k + k]) {
            nemu = 1;
            break;
          }
        }
        if (nemu) break;
      }
      if (nemu) {
        cnt += n - r;
        break;
      }
    }
  }
  cout << cnt << '\n';
  return 0;
}

```