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;
bool test(long long len) {
  long long untill = (1LL << len);
  for (long long val = 0; val < untill; val++) {
    bool pos = false;
    for (long long x = 0; x < len; x++) {
      for (long long k = 1; k < len; k++) {
        long long a = (1LL << x), b = (1LL << (x + k)),
                  c = (1LL << (x + 2 * k));
        if (x + 2 * k > len) continue;
        if ((bool)(val & a) == (bool)(val & b) &&
            (bool)(val & a) == (bool)(val & c)) {
          pos = true;
          break;
        }
      }
    }
    if (!pos) return false;
  }
  cerr << "[DEBUG] "
       << "untill - 1"
       << " = " << untill - 1 << endl;
  ;
  return true;
}
void brute() {
  for (long long i = (long long)(1); i <= (long long)(14); i++) {
    cerr << i << " = " << test(i) << '\n';
  }
}
string s;
int main() {
  string s;
  cin >> s;
  long long ans = 0;
  set<pair<long long, long long>> p, rp;
  for (long long x = 0; x < s.size(); x++) {
    long long untill = min((long long)s.size() - 1LL, x + 20);
    for (long long k = 1; k <= 10; k++) {
      long long a = x, b = x + k, c = x + 2 * k;
      if (c > untill) break;
      if (s[a] == s[b] && s[a] == s[c]) {
        p.insert({x, c});
        rp.insert({c, x});
        break;
      }
    }
  }
  for (long long x = 0; x < s.size(); x++) {
    if (!rp.empty()) {
      ans += (s.size() - rp.begin()->first);
    }
    if (!p.empty() && p.begin()->first <= x) {
      auto cur = *p.begin();
      rp.erase({cur.second, cur.first});
      p.erase(p.begin());
    }
  }
  cout << ans;
}