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 X, class Y>
bool minimize(X &x, const Y &y) {
  X eps = 1e-9;
  if (x > y + eps) {
    x = y;
    return true;
  } else
    return false;
}
template <class X, class Y>
bool maximize(X &x, const Y &y) {
  X eps = 1e-9;
  if (x + eps < y) {
    x = y;
    return true;
  } else
    return false;
}
template <class T>
T Abs(const T &x) {
  return (x < 0 ? -x : x);
}
char s[300300];
int n;
bool ok(int l, int r) {
  for (int k = (1), _b = ((r - l + 1) / 2); k <= _b; k++)
    for (int i = (l + k), _b = (r - k); i <= _b; i++)
      if (s[i] == s[i - k] && s[i] == s[i + k]) return true;
  return false;
}
void process(void) {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  long long res = 0;
  for (int i = (1), _b = (n); i <= _b; i++) {
    if (i + 11 <= n) res += n - (i + 11) + 1;
    for (int j = (i), _b = (n); j <= _b; j++) {
      if (j > i + 10) break;
      if (ok(i, j)) res++;
    }
  }
  cout << res << endl;
}
int main(void) {
  process();
  return 0;
}