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;
const int N = int(1e6) + 10;
const int M = int(7e5) + 10;
const int inf = int(1e9);
const int mod = int(1e9) + 7;
const long double eps = 1e-15;
void solve() {
  string s;
  cin >> s;
  long long ans = 0;
  for (int i = 0; i < s.length(); ++i) {
    vector<int> z, m;
    for (int j = i; j < s.length(); ++j) {
      bool b = false;
      if (s[j] == '0') {
        for (int l = 0; l < z.size(); ++l) {
          if ((z[l] + j) % 2 == 0 && s[(z[l] + j) / 2] == '0') {
            b = true;
            break;
          }
        }
        z.push_back(j);
      } else {
        for (int l = 0; l < m.size(); ++l) {
          if ((m[l] + j) % 2 == 0 && s[(m[l] + j) / 2] == '1') {
            b = true;
            break;
          }
        }
        m.push_back(j);
      }
      if (b) {
        ans += s.length() - j;
        break;
      }
    }
  }
  printf("%lld\n", ans);
}
int main() {
  solve();
  return 0;
}