#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
char str[maxn];
int d[maxn];
int main() {
  scanf("%s", str + 1);
  int n = strlen(str + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; i + j * 2 <= n; j++) {
      if (str[i] == str[i + j] && str[i + j] == str[i + 2 * j]) {
        d[i] = i + j * 2;
        break;
      }
    }
  long long ans = 0, r = 1e9;
  for (int i = n; i >= 1; i--) {
    if (d[i]) {
      r = min(r, (long long)d[i]);
    }
    if (r != 1e9) ans += n - r + 1;
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int maxn = 3e5 + 10;
char str[maxn];
int d[maxn];
int main() {
  scanf("%s", str + 1);
  int n = strlen(str + 1);
  for (int i = 1; i <= n; i++)
    for (int j = 1; i + j * 2 <= n; j++) {
      if (str[i] == str[i + j] && str[i + j] == str[i + 2 * j]) {
        d[i] = i + j * 2;
        break;
      }
    }
  long long ans = 0, r = 1e9;
  for (int i = n; i >= 1; i--) {
    if (d[i]) {
      r = min(r, (long long)d[i]);
    }
    if (r != 1e9) ans += n - r + 1;
  }
  cout << ans;
  return 0;
}

```