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;
string make(int n, int cur) {
  string result = to_string(cur);
  int left = n - result.size();
  string zero(left, '0');
  return zero + result;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string in;
  cin >> in;
  vector<pair<int, int> > all;
  for (int i = 0; i < in.size(); i++) {
    pair<int, int> cur = {i, -1};
    for (int k = 1; k <= 100; k++) {
      if (i + 2 * k >= in.size()) {
        break;
      }
      if (in[i] == in[i + k] && in[i] == in[i + 2 * k]) {
        cur = {i, i + 2 * k};
        break;
      }
    }
    all.push_back(cur);
  }
  int small = all.back().second;
  long long result = 0;
  for (int i = all.size() - 1; i >= 0; i--) {
    if (small == -1) {
      small = all[i].second;
    }
    if (all[i].second != -1 && all[i].second < small) {
      small = all[i].second;
    }
    if (small != -1) {
      result += in.size() - small;
    }
  }
  cout << result;
}