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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string p;
  cin >> p;
  string s;
  s = "$";
  for (auto k : p) s += k;
  set<long long int> v[2];
  long long int n = p.size();
  long long int arr[n + 1];
  for (long long int i = 1; i <= n; i++) {
    long long int pos = s[i] - '0';
    v[pos].insert(i);
    arr[i] = -1;
  }
  for (long long int i = 1; i <= n; i++) {
    long long int pos = s[i] - '0';
    while (v[pos].size() != 0 && (*v[pos].begin() <= i))
      v[pos].erase(*v[pos].begin());
    for (auto idx : v[pos]) {
      long long int x = i;
      long long int k = idx - i;
      long long int xk = x + k;
      long long int x2k = x + 2 * k;
      if (x2k > n) {
        break;
      }
      if (s[x] == s[xk] && s[xk] == s[x2k]) {
        arr[i] = x2k;
        break;
      }
    }
  }
  long long int mn = -1, ans = 0;
  for (long long int i = n; i > 0; i--) {
    if (arr[i] == -1 && mn == -1) {
      continue;
    } else if (mn == -1)
      mn = arr[i];
    else if (arr[i] != -1)
      mn = min(mn, arr[i]);
    long long int cnt = n - mn + 1;
    ans += cnt;
  }
  cout << ans;
  return 0;
}