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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int nmax = 2000 * 1000;
const int inf = 2000 * 1000 * 1000;
const long long infl = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 1000ll;
const int mod = 1000 * 1000 * 1000 + 7;
const long double pi = acos(-1.0);
template <typename T1, typename T2>
ostream& operator<<(ostream& cout, const pair<T1, T2>& a) {
  return (cout << "(" << a.first << "; " << a.second << ")");
}
template <typename T>
ostream& operator<<(ostream& cout, const vector<T>& a) {
  if (!a.size()) {
    return (cout << "()");
  }
  cout << "(" << a[0];
  for (int i = 1; i < (int)a.size(); i++) {
    cout << "; " << a[i];
  }
  return (cout << ")");
}
template <typename T>
ostream& operator<<(ostream& cout, const set<T>& a) {
  if (!a.size()) {
    return (cout << "{}");
  }
  auto it = a.begin();
  cout << "{" << *it;
  ++it;
  for (; it != a.end(); ++it) {
    cout << "; " << *it;
  }
  return (cout << "}");
}
string s;
int ans[nmax];
int main() {
  getline(cin, s);
  int n = s.length();
  for (int i = 0; i < n; i++) {
    ans[i] = -1;
    for (int j = 1; j < 600; j++) {
      if (i + 2 * j >= n) break;
      if (s[i] == s[i + j] && s[i + j] == s[i + 2 * j]) {
        ans[i] = j;
        break;
      }
    }
  }
  long long answer = 0;
  long long cur = -1;
  for (int i = n - 1; i >= 0; i--) {
    long long r = (ans[i] == -1 ? -1 : i + 2 * ans[i]);
    if (cur == -1 && r == -1) continue;
    if (cur == -1) cur = r;
    if (cur != -1 && r != -1) cur = min(cur, r);
    if (cur == -1) continue;
    answer += n - cur;
  }
  cout << answer << endl;
  return 0;
}