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;
void run() {
  iostream::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  string s;
  cin >> s;
  long long int n = s.length();
  map<long long int, long long int> m[2];
  for (long long i = (0); i < (n); ++i) {
    if (s[i] == '0')
      m[0][i] = 1;
    else
      m[1][i] = 1;
  }
  long long int ans = 0;
  long long int prev[n + 1];
  prev[n] = n;
  for (long long int i = n - 1; i >= 0; i--) {
    prev[i] = prev[i + 1];
    for (long long int k = 1; i + 2 * k < prev[i]; k++) {
      if (s[i + k] == s[i] && s[i + 2 * k] == s[i]) {
        prev[i] = (i + 2 * k);
      }
    }
  }
  for (long long i = (0); i < (n); ++i) {
    ans += (n - prev[i]);
  }
  cout << ans;
}
int main() { run(); }