#include <bits/stdc++.h>
using namespace std;
char s[300005];
int n;
bool ok(int l, int r) {
  int len = r - l + 1;
  for (int x = 1; 2 * x + 1 <= len; x++)
    for (int k = l; k + 2 * x <= r; k++)
      if (s[k] == s[k + x] && s[k + x] == s[k + 2 * x]) return 1;
  return 0;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    for (int j = 1; j < 10; j++)
      if (i + j - 1 <= n) cnt += !ok(i, i + j - 1);
    ans += n - i + 1 - cnt;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
char s[300005];
int n;
bool ok(int l, int r) {
  int len = r - l + 1;
  for (int x = 1; 2 * x + 1 <= len; x++)
    for (int k = l; k + 2 * x <= r; k++)
      if (s[k] == s[k + x] && s[k + x] == s[k + 2 * x]) return 1;
  return 0;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    for (int j = 1; j < 10; j++)
      if (i + j - 1 <= n) cnt += !ok(i, i + j - 1);
    ans += n - i + 1 - cnt;
  }
  cout << ans << endl;
  return 0;
}

```