#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int n = s.size();
  int l[n];
  for (int i = 0; i < n; ++i) {
    l[i] = n;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= min(n - i - 1, i); ++j) {
      if (s[i - j] == s[i] && s[i] == s[i + j]) {
        l[i - j] = min(l[i - j], i + j);
        break;
      }
    }
  }
  set<pair<int, int> > h;
  for (int i = 0; i < n; ++i) {
    h.insert({l[i], i});
  }
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += n - ((*h.begin()).first);
    h.erase({l[i], i});
  }
  cout << sum;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  string s;
  cin >> s;
  int n = s.size();
  int l[n];
  for (int i = 0; i < n; ++i) {
    l[i] = n;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= min(n - i - 1, i); ++j) {
      if (s[i - j] == s[i] && s[i] == s[i + j]) {
        l[i - j] = min(l[i - j], i + j);
        break;
      }
    }
  }
  set<pair<int, int> > h;
  for (int i = 0; i < n; ++i) {
    h.insert({l[i], i});
  }
  long long sum = 0;
  for (int i = 0; i < n; ++i) {
    sum += n - ((*h.begin()).first);
    h.erase({l[i], i});
  }
  cout << sum;
  return 0;
}

```