#include <bits/stdc++.h>
using namespace std;
long long a[600008], vs[500000];
string s, s1;
map<long long, long long> mp;
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  };
  long long i, j, k = 0, l, n, m, t, h, cnt = 0, curr, mx, mn, mod = 1e9 + 7;
  cin >> s;
  n = s.size();
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= 10; j++) {
      if (i - 2 * j > 0) {
        if (s[i - 1] == s[i - j - 1] && s[i - 1] == s[i - 2 * j - 1]) {
          vs[i] = 2 * j + 1;
          break;
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    mn = INT_MAX;
    for (j = i; j > max(k, i - 20); j--) {
      if (vs[j]) mn = min(mn, vs[j] + i - j);
    }
    if (mn != INT_MAX) cnt += i - mn + 1;
  }
  cout << cnt << endl;
  return 0;
}

### Prompt

Generate a cpp solution 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;
long long a[600008], vs[500000];
string s, s1;
map<long long, long long> mp;
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  };
  long long i, j, k = 0, l, n, m, t, h, cnt = 0, curr, mx, mn, mod = 1e9 + 7;
  cin >> s;
  n = s.size();
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= 10; j++) {
      if (i - 2 * j > 0) {
        if (s[i - 1] == s[i - j - 1] && s[i - 1] == s[i - 2 * j - 1]) {
          vs[i] = 2 * j + 1;
          break;
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    mn = INT_MAX;
    for (j = i; j > max(k, i - 20); j--) {
      if (vs[j]) mn = min(mn, vs[j] + i - j);
    }
    if (mn != INT_MAX) cnt += i - mn + 1;
  }
  cout << cnt << endl;
  return 0;
}

```