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 MAXN = 300005;
const int INF = 1 << 30;
char s[MAXN];
int r[MAXN];
int n;
vector<int> pos[2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> (s + 1), n = strlen(s + 1);
  for (int i = 1; i <= n; i++) pos[s[i] - '0'].push_back(i);
  for (int id = 0; id < 2; id++) {
    for (int i = 0; i < pos[id].size(); i++) {
      int x = pos[id][i];
      r[x] = INF;
      for (int j = i + 1; j < pos[id].size(); j++) {
        int k = pos[id][j] - pos[id][i];
        int y = x + 2 * k;
        if (y <= n && s[y] - '0' == id) {
          r[x] = y;
          break;
        }
      }
    }
  }
  for (int i = n - 1; i >= 1; i--)
    if (r[i] > r[i + 1]) r[i] = r[i + 1];
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    if (r[i] != INF) res += n - r[i] + 1;
  }
  cout << res << endl;
  return 0;
}