#include <bits/stdc++.h>
using namespace std;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  string inp;
  cin >> inp;
  long long n = inp.size();
  long long dp[n];
  memset(dp, 0, sizeof dp);
  long long ans = 0;
  for (long long int l = 0; l < n; l++) {
    for (long long int k = 1; k < 21; k++) {
      if (l + 2 * k < n && inp[l] == inp[l + k] && inp[l] == inp[l + 2 * k]) {
        dp[l] = (n - l - 2 * k);
        break;
      }
    }
  }
  for (long long i = n - 2; i >= 0; i--) {
    dp[i] = max(dp[i], dp[i + 1]);
    ans += dp[i];
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  string inp;
  cin >> inp;
  long long n = inp.size();
  long long dp[n];
  memset(dp, 0, sizeof dp);
  long long ans = 0;
  for (long long int l = 0; l < n; l++) {
    for (long long int k = 1; k < 21; k++) {
      if (l + 2 * k < n && inp[l] == inp[l + k] && inp[l] == inp[l + 2 * k]) {
        dp[l] = (n - l - 2 * k);
        break;
      }
    }
  }
  for (long long i = n - 2; i >= 0; i--) {
    dp[i] = max(dp[i], dp[i + 1]);
    ans += dp[i];
  }
  cout << ans << '\n';
  return 0;
}

```