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 maxp = 22;
const long double EPS = 1e-18;
const long long INF = 1e18;
const int MOD = 1e9 + 7;
const int N = 2e5 + 1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = s.size(), i;
  long long ans = 0;
  vector<vector<int> > idx(2);
  int p[2] = {1, 1};
  for (i = 0; i < n; i++) idx[s[i] - '0'].push_back(i);
  vector<pair<int, int> > seg;
  for (i = 0; i < n - 2; i++) {
    int curr = s[i] - '0';
    int pos = p[curr];
    p[curr]++;
    int r = n;
    while (pos + 1 < idx[curr].size()) {
      int k = idx[curr][pos] - i;
      if (idx[curr][pos] + k < n && s[idx[curr][pos] + k] == s[i]) {
        r = idx[curr][pos] + k;
        break;
      }
      pos++;
    }
    if (r != n) {
      while (seg.size() && seg.back().second >= r) seg.pop_back();
      seg.push_back({i, r});
    }
  }
  int l = -1;
  for (auto x : seg) {
    ans += (x.first - l) * (n - x.second);
    l = x.first;
  }
  cout << ans;
}