#include <bits/stdc++.h>
using namespace std;
signed main() {
  string str;
  cin >> str;
  long long n = str.length();
  vector<long long> vec(n, INT_MAX);
  for (long long i = 0; i < n; ++i) {
    for (long long j = i + 2; (j <= (i + 10) && (j < n)); j += 2) {
      if ((str[i] == str[j]) && (str[i] == str[(i + j) / 2])) {
        vec[i] = j;
        break;
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    long long rit = INT_MAX;
    for (long long j = i; (j <= i + 10) && (j < n); ++j) {
      rit = min(rit, vec[j]);
    }
    if (rit != INT_MAX) {
      ans += (n - rit);
    }
  }
  cout << ans << "\n";
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 str;
  cin >> str;
  long long n = str.length();
  vector<long long> vec(n, INT_MAX);
  for (long long i = 0; i < n; ++i) {
    for (long long j = i + 2; (j <= (i + 10) && (j < n)); j += 2) {
      if ((str[i] == str[j]) && (str[i] == str[(i + j) / 2])) {
        vec[i] = j;
        break;
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    long long rit = INT_MAX;
    for (long long j = i; (j <= i + 10) && (j < n); ++j) {
      rit = min(rit, vec[j]);
    }
    if (rit != INT_MAX) {
      ans += (n - rit);
    }
  }
  cout << ans << "\n";
}

```