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;
int n, m, k, y, z, l, i, j, x, r, x2, y2;
int a[300500], b[100500];
bool good(int x) {
  int p = 0;
  for (int i = 0; i < n; i++) {
    if ((a[i] + x) % m < a[i] && (a[i] + x) % m >= p) {
      continue;
    }
    if (p <= a[i] + x && a[i] <= p) {
      continue;
    }
    if (a[i] > p) {
      p = a[i];
      continue;
    }
    return false;
  }
  return true;
}
int main() {
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    a[i] = n + 2;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 1; j < 1000; j++) {
      if (i - 2 * j < 0) {
        break;
      }
      if (s[i] == s[i - j] && s[i - j] == s[i - 2 * j]) {
        a[i - 2 * j] = min(a[i - 2 * j], i);
      }
    }
  }
  for (int i = n - 2; i >= 0; i--) {
    a[i] = min(a[i], a[i + 1]);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += max(0, n - a[i]);
  }
  cout << ans << endl;
  return 0;
}