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>
const double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
using namespace std;
const int MOD = 1e9 + 7;
double gcd(double a, double b) { return a < 0.01 ? b : gcd(fmod(b, a), a); }
template <typename T>
T mymax(T a, T b) {
  return (a > b) ? a : b;
}
template <typename T>
T mymax(T a, T b, T c) {
  return mymax(a, mymax(b, c));
}
template <typename T>
T mymin(T a, T b) {
  return (a < b) ? a : b;
}
template <typename T>
T mymin(T a, T b, T c) {
  return mymin(a, mymin(b, c));
}
template <typename T>
T power(T x, T y) {
  T res = 1;
  x = x % MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return res;
}
template <typename T>
void swap(T *x, T *y) {
  T temp;
  temp = *y;
  *y = *x;
  *x = temp;
}
template <typename T>
T mod(T a) {
  if (a > 0)
    return a;
  else
    return -a;
}
template <typename T>
T logg(T a) {
  T x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int ans = 0;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  vector<int> p(n + 1, n);
  for (int i = n - 1; i >= 0; i--) {
    p[i] = p[i + 1];
    for (int k = 1; i + 2 * k <= p[i]; k++) {
      if (s[i] == s[i + k] && s[i] == s[i + 2 * k]) {
        p[i] = i + 2 * k;
      }
    }
    ans += (n - p[i]);
  }
  cout << ans << endl;
  return 0;
}