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 int mod = 1e9 + 7, test;
long long int max(long long int a, long long int b) { return a > b ? a : b; }
long long int min(long long int a, long long int b) { return a > b ? b : a; }
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  string s;
  cin >> s;
  long long int n = s.length();
  long long int cnt = 0;
  for (long long int i = 0; i < (n); i++) {
    long long int r = n, en = i + 10;
    for (long long int x = i; x < en; ++x) {
      for (long long int k = 1; x + 2 * k < en; ++k) {
        if (s[x] == s[x + k] && s[x + k] == s[x + 2 * k]) {
          r = min(r, x + 2 * k);
        }
      }
    }
    cnt += (n - r);
  }
  cout << cnt << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t = 1;
  for (long long int i = 0; i < (t); i++) {
    solve();
  }
  return 0;
}