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;
int main() {
  unordered_set<string> bad;
  for (int L = 3; L <= 8; L++) {
    for (int i = 0; i < (1 << L); i++) {
      int c = i;
      string v;
      for (int j = 0; j < L; j++) {
        if (c % 2)
          v += '1';
        else
          v += '0';
        c /= 2;
      }
      bool hast = false;
      for (int x = 0; x < L; x++)
        for (int k = 1; !hast && x + 2 * k < L; k++)
          if (v[x] == v[x + k] && v[x] == v[x + 2 * k]) {
            hast = true;
            break;
          }
      if (!hast) {
        bad.insert(v);
      }
    }
  }
  long long res = 0;
  string s;
  cin >> s;
  int n = s.length();
  for (int l = 0; l < n; l++) {
    for (int t = 3; l + t - 1 < n && t <= 8; t++) {
      if (bad.find(s.substr(l, t)) == bad.end()) {
        res++;
      }
    }
    res += max(0LL, (long long)n - (l + 8));
  }
  cout << res << endl;
  return 0;
}