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 = 3e5 + 10;
inline int read() {
  register int x = 0, f = 0;
  register char c = getchar();
  while (c < '0' || c > '9') f |= c == '-', c = getchar();
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return f ? -x : x;
}
int n;
char s[N];
namespace Check {
bool check(int S) {
  for (int i = 0; i < 4; ++i)
    if ((S >> i) & 1)
      s[i + 2] = '1';
    else
      s[i + 2] = '0';
  for (int i = 4; i < 8; ++i)
    if ((S >> i) & 1)
      s[i + 3] = '1';
    else
      s[i + 3] = '0';
  for (int i = 1; i <= 11; ++i)
    for (int j = 1; i + j * 2 <= 11; ++j)
      if (s[i] == s[i + j] && s[i] == s[i + j * 2]) return false;
  return true;
}
int main() {
  s[1] = s[6] = s[11] = '0';
  for (int i = 0; i < (1 << 8); ++i)
    if (check(i)) puts("RQ");
  return 0;
}
}  // namespace Check
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  long long ans = 0;
  int L = 0;
  for (int r = 1; r <= n; ++r) {
    for (int l = r - 1; l >= 1; l--)
      if ((l + r) % 2 == 0 && s[l] == s[r] && s[(l + r) / 2] == s[r]) {
        L = max(l, L);
        break;
      }
    ans += L;
  }
  cout << ans << '\n';
  return 0;
}