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 = 300003;
char s[N];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  vector<pair<int, int> > seg;
  vector<int> idx[2];
  for (int i = 0; i < n; ++i) idx[s[i] - '0'].push_back(i);
  int L = sqrt(n) * 2 + 2;
  for (int it = 0; it < 2; ++it) {
    auto &v = idx[it];
    for (int i = 0; i < v.size(); ++i)
      for (int j = i + 1; j + 1 < v.size(); ++j) {
        const int at = v[j];
        const int k = at - v[i];
        if (at + k < n && s[at + k] == it + '0') {
          seg.push_back({at - k, at + k});
          break;
        }
        if (j - i > L) break;
      }
  }
  sort(seg.begin(), seg.end());
  long long res = 0;
  int j = (signed)seg.size() - 1;
  int mn = n;
  for (int i = n - 1; i >= 0; --i) {
    while (j >= 0 && seg[j].first >= i) {
      mn = min(mn, seg[j].second);
      --j;
    }
    res += n - mn;
  }
  printf("%lld\n", res);
  return 0;
}