#include <bits/stdc++.h>
using namespace std;
int n;
string s;
int x[300002];
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) {
    x[i] = -1;
    for (int j = i + 1; i + (j - i) * 2 < n; j++)
      if (s[i] == s[j] && s[j] == s[i + (j - i) * 2]) {
        x[i] = i + (j - i) * 2;
        break;
      }
  }
  int best = -1;
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (x[i] != -1 && (x[i] < best || best == -1)) best = x[i];
    if (best != -1) ans += n - best;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Generate 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;
int n;
string s;
int x[300002];
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) {
    x[i] = -1;
    for (int j = i + 1; i + (j - i) * 2 < n; j++)
      if (s[i] == s[j] && s[j] == s[i + (j - i) * 2]) {
        x[i] = i + (j - i) * 2;
        break;
      }
  }
  int best = -1;
  long long ans = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (x[i] != -1 && (x[i] < best || best == -1)) best = x[i];
    if (best != -1) ans += n - best;
  }
  cout << ans << "\n";
  return 0;
}

```