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;
const int N = 1e6;
int nx[2][N + 2], aa[N + 2], mn[N + 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  nx[0][n] = n;
  nx[1][n] = n;
  for (int i = n - 1; i >= 0; i--) {
    nx[0][i] = nx[0][i + 1];
    nx[1][i] = nx[1][i + 1];
    if (s[i] == '1')
      nx[1][i] = i;
    else
      nx[0][i] = i;
  }
  for (int i = 0; i < n; i++) {
    int c = s[i] - '0';
    aa[i] = n;
    int first = nx[c][i + 1];
    while (true) {
      if (first == n) break;
      int second = first + first - i;
      if (second >= n) break;
      first = nx[c][first + 1];
      if (s[second] != s[i]) continue;
      aa[i] = second;
      break;
    }
  }
  mn[n] = n;
  for (int i = n - 1; i >= 0; i--) mn[i] = min(mn[i + 1], aa[i]);
  long long ans = 0;
  for (int i = 0; i < n; i++) ans += (n - mn[i]) * 1LL;
  cout << ans << endl;
  return 0;
}