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;
void cre(vector<vector<int> > &vv, int n, int idx, vector<int> v, int m) {
  if (m == 0) {
    vv.push_back(v);
    return;
  }
  if (idx == n) {
    if (m == 0) vv.push_back(v);
    return;
  }
  v[idx] = 1;
  cre(vv, n, idx + 1, v, m - 1);
  v[idx] = 0;
  cre(vv, n, idx + 1, v, m);
}
int bsearch(vector<int> &v, int l, int r, int val) {
  if (l >= r) return l;
  int mid = (l + r) / 2;
  if (v[mid] == val) return mid;
  if (v[mid] < val) {
    if (v[mid + 1] > val) return mid;
    return bsearch(v, mid + 1, r, val);
  }
  return bsearch(v, l, mid, val);
}
int bsearch2(vector<int> &idx, vector<int> &v, int l, int r, int val) {
  if (l >= r) return l;
  int mid = (l + r) / 2;
  if (v[idx[mid]] <= val) {
    if (v[idx[mid + 1]] > val) return mid;
    return bsearch2(idx, v, mid + 1, r, val);
  }
  return bsearch2(idx, v, l, mid, val);
}
void crefac(vector<vector<int> > &vv, int n, int idx, vector<int> v,
            set<int> s) {
  if (idx == n) {
    vv.push_back(v);
    return;
  }
  for (auto it = s.begin(); it != s.end(); it++) {
    int x = *it;
    v[idx] = x;
    set<int> xx = s;
    xx.erase(x);
    crefac(vv, n, idx + 1, v, xx);
  }
}
int findx(string &s, int start, int stop) {
  for (int i = start + 2; i <= stop; i++) {
    int k = 1;
    while (i - 2 * k >= start) {
      if (s[i] == s[i - k] && s[i] == s[i - 2 * k]) return i;
      k += 1;
    }
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  long long ans = 0;
  for (int i = 0; i < s.size(); i++) {
    int x = findx(s, i, min(i + 8, (int)s.size() - 1));
    if (x == -1) continue;
    ans += s.size() - x;
  }
  cout << ans << endl;
  return 0;
}