#include <bits/stdc++.h>
using namespace std;
signed main() {
  string s;
  cin >> s;
  long long n = s.size();
  long long an = 0;
  for (long long i = 0; i < n; i++) {
    long long r = INT_MAX;
    for (long long j = i; j <= min((long long)(i + 9), n - 1); j++) {
      for (long long k = 1; k <= 7; k++) {
        if (j + 2 * k < n) {
          if (s[j] == s[j + k] and s[j] == s[j + 2 * k]) {
            r = min(r, j + 2 * k);
          }
        }
      }
    }
    if (r != INT_MAX) an = an + n - r;
  }
  cout << an << endl;
}

### Prompt

Your challenge is to write 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;
signed main() {
  string s;
  cin >> s;
  long long n = s.size();
  long long an = 0;
  for (long long i = 0; i < n; i++) {
    long long r = INT_MAX;
    for (long long j = i; j <= min((long long)(i + 9), n - 1); j++) {
      for (long long k = 1; k <= 7; k++) {
        if (j + 2 * k < n) {
          if (s[j] == s[j + k] and s[j] == s[j + 2 * k]) {
            r = min(r, j + 2 * k);
          }
        }
      }
    }
    if (r != INT_MAX) an = an + n - r;
  }
  cout << an << endl;
}

```