#include <bits/stdc++.h>
using namespace std;
const int MaxN = 3e5;
char s[MaxN + 5];
long long Solve(int l, int r, int cas) {
  if (cas == 5) return 0LL;
  long long ans = 0;
  for (int i = l + 2 * cas; i <= r; i++) {
    if (s[i] == s[i - cas] && s[i] == s[i - 2 * cas]) {
      ans = (long long)(r - i + 1) * (i - 2 * cas - l + 1);
      ans += Solve(l, i - 1, cas + 1);
      ans += Solve(i - 2 * cas + 1, r, cas);
      return ans;
    }
  }
  return Solve(l, r, cas + 1);
}
int main() {
  int n;
  long long ans = 0;
  scanf(" %s", s + 1);
  n = strlen(s + 1);
  ans = Solve(1, n, 1);
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
char s[MaxN + 5];
long long Solve(int l, int r, int cas) {
  if (cas == 5) return 0LL;
  long long ans = 0;
  for (int i = l + 2 * cas; i <= r; i++) {
    if (s[i] == s[i - cas] && s[i] == s[i - 2 * cas]) {
      ans = (long long)(r - i + 1) * (i - 2 * cas - l + 1);
      ans += Solve(l, i - 1, cas + 1);
      ans += Solve(i - 2 * cas + 1, r, cas);
      return ans;
    }
  }
  return Solve(l, r, cas + 1);
}
int main() {
  int n;
  long long ans = 0;
  scanf(" %s", s + 1);
  n = strlen(s + 1);
  ans = Solve(1, n, 1);
  printf("%I64d\n", ans);
  return 0;
}

```