#include <bits/stdc++.h>
using namespace std;
int n;
char s[300005];
long long ans;
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int l = 0; l < n; l++) {
    int r = l;
    while (1) {
      bool check = 1;
      for (int i = l + 1; i < r; i++) {
        int d = r - i;
        if (i - d >= l && s[r] == s[i] && s[i] == s[i - d]) {
          check = 0;
          break;
        }
      }
      if (!check) break;
      r++;
    }
    if (r < n) ans += (long long)(n - r);
  }
  cout << ans << endl;
  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;
int n;
char s[300005];
long long ans;
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int l = 0; l < n; l++) {
    int r = l;
    while (1) {
      bool check = 1;
      for (int i = l + 1; i < r; i++) {
        int d = r - i;
        if (i - d >= l && s[r] == s[i] && s[i] == s[i - d]) {
          check = 0;
          break;
        }
      }
      if (!check) break;
      r++;
    }
    if (r < n) ans += (long long)(n - r);
  }
  cout << ans << endl;
  return 0;
}

```