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;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const long long INF = 1e18;
string S;
long long ans = 0;
bool g(string S) {
  for (int i = 0; i < S.size(); i++) {
    for (int j = 1; i + 2 * j < S.size(); j++) {
      if (S[i] == S[i + j] and S[i] == S[i + 2 * j]) return true;
    }
  }
  return false;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> S;
  long long r = 1;
  for (long long l = 0; l < S.size(); l++) {
    if (r == S.size() + 1) break;
    while (r != (S.size() + 1) and !g(S.substr(l, r - l))) r++;
    ans += S.size() - r + 1;
  }
  cout << ans << "\n";
  return 0;
}