#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long inf18 = 1e18 + 7;
const int N = 3e5 + 7;
int c[N], l[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  string s;
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) l[i] = -1;
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j < n; ++j) {
      int zn = (j << 1);
      if (zn + i >= n) break;
      if (s[i] == s[i + j] && s[i] == s[zn + i]) {
        c[i] = j;
        break;
      }
    }
    if (c[i]) l[c[i] + c[i] + i] = max(l[c[i] + c[i] + i], i);
  }
  for (int i = 1; i < n; ++i) l[i] = max(l[i], l[i - 1]);
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += l[i] + 1;
  cout << ans;
}

### Prompt

Please provide a Cpp coded solution 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;
const int inf = 1e9 + 7;
const long long inf18 = 1e18 + 7;
const int N = 3e5 + 7;
int c[N], l[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  string s;
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) l[i] = -1;
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j < n; ++j) {
      int zn = (j << 1);
      if (zn + i >= n) break;
      if (s[i] == s[i + j] && s[i] == s[zn + i]) {
        c[i] = j;
        break;
      }
    }
    if (c[i]) l[c[i] + c[i] + i] = max(l[c[i] + c[i] + i], i);
  }
  for (int i = 1; i < n; ++i) l[i] = max(l[i], l[i - 1]);
  long long ans = 0;
  for (int i = 0; i < n; ++i) ans += l[i] + 1;
  cout << ans;
}

```