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;
using ll = long long int;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vint = vector<int>;
class solver {
 public:
  void solve(string &s) {
    ll len = s.size();
    ll ans = len * (len + 1) / 2;
    ll rig = s.size();
    for (int pos = s.size() - 1; pos >= 0; pos--) {
      for (int new_rig = pos + 2; new_rig < rig; new_rig++) {
        if ((new_rig - pos) % 2 == 0) {
          if (s[pos] == s[new_rig] && s[pos] == s[(pos + new_rig) / 2]) {
            rig = new_rig;
            break;
          }
        }
      }
      ans -= s.size() - rig;
    }
    cout << (len * (len + 1) / 2 - ans) << endl;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s;
  while (cin >> s) {
    solver ss;
    ss.solve(s);
  }
}