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;
long long p = 1000000007ll;
int main(void) {
  set<string> has;
  for (long long n = (0); n < (10); ++n) {
    for (long long i = (0); i < ((1 << n)); ++i) {
      string s;
      for (long long j = (0); j < (n); ++j) s.push_back('0' + ((i >> j) & 1));
      bool flag = false;
      for (long long k = (0); k < (n); ++k) {
        for (long long l = (1); l < (n); ++l) {
          if (k + l * 2 >= n) break;
          if (s[k] == s[k + l] && s[k] == s[k + l * 2]) flag = true;
        }
      }
      if (flag) has.insert(s);
    }
  }
  string is;
  cin >> is;
  long long answer = is.size();
  answer = answer * (answer - 1) / 2;
  long long l = 0, r = 1;
  while (r <= is.size()) {
    if (has.find(string(is.begin() + l, is.begin() + r)) == has.end()) {
      answer -= r - l - 1;
      ++r;
    } else {
      ++l;
    }
  }
  cout << answer << endl;
  return 0;
}