#include <bits/stdc++.h>
using namespace std;
const int PRIME = 1e9 + 7;
const int INF = 1e8;
const int MAX_N = 3e5 + 6;
long long ans = 0;
int n, m, k;
string s;
int a[MAX_N];
int nxtj[MAX_N];
int nxtz[MAX_N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  int zer = INF, jed = INF;
  for (int i = (0); i <= (n); ++i) a[i] = n;
  for (int i = (int)(n - 1); i >= (int)(0); --i) {
    nxtj[i] = jed;
    nxtz[i] = zer;
    if (s[i] == '0') {
      zer = i;
    } else {
      jed = i;
    }
  }
  for (int i = (int)(n - 2); i >= (int)(0); --i) {
    int k;
    bool b = s[i] == '1';
    if (b)
      k = nxtj[i] - i;
    else
      k = nxtz[i] - i;
    a[i] = a[i + 1];
    while (i + 2 * k < a[i]) {
      if (s[i + 2 * k] == s[i]) {
        a[i] = i + 2 * k;
      }
      if (b)
        k = nxtj[i + k] - i;
      else
        k = nxtz[i + k] - i;
    }
  }
  for (int i = (0); i < (n); ++i) {
    if (a[i] == n) break;
    ans += n - a[i];
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int PRIME = 1e9 + 7;
const int INF = 1e8;
const int MAX_N = 3e5 + 6;
long long ans = 0;
int n, m, k;
string s;
int a[MAX_N];
int nxtj[MAX_N];
int nxtz[MAX_N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  int zer = INF, jed = INF;
  for (int i = (0); i <= (n); ++i) a[i] = n;
  for (int i = (int)(n - 1); i >= (int)(0); --i) {
    nxtj[i] = jed;
    nxtz[i] = zer;
    if (s[i] == '0') {
      zer = i;
    } else {
      jed = i;
    }
  }
  for (int i = (int)(n - 2); i >= (int)(0); --i) {
    int k;
    bool b = s[i] == '1';
    if (b)
      k = nxtj[i] - i;
    else
      k = nxtz[i] - i;
    a[i] = a[i + 1];
    while (i + 2 * k < a[i]) {
      if (s[i + 2 * k] == s[i]) {
        a[i] = i + 2 * k;
      }
      if (b)
        k = nxtj[i + k] - i;
      else
        k = nxtz[i + k] - i;
    }
  }
  for (int i = (0); i < (n); ++i) {
    if (a[i] == n) break;
    ans += n - a[i];
  }
  cout << ans;
  return 0;
}

```