#include <bits/stdc++.h>
using namespace std;
using lli = long long;
const int maxTrial = 10;
int cnt;
int x[maxTrial];
set<string> Set;
string s;
inline bool Check(int i) {
  for (int j = i - 2; j >= 0; j -= 2)
    if (x[i] == x[j] && x[i] == x[(i + j) / 2]) return false;
  return true;
}
void Print(int i) {
  string s;
  s.resize(i + 1);
  for (int j = 0; j <= i; ++j) s[j] = x[j] + '0';
  Set.insert(s);
}
void Attempt(int i) {
  for (x[i] = 0; x[i] <= 1; ++x[i])
    if (Check(i)) {
      Print(i);
      Attempt(i + 1);
    }
}
void Generate() {
  cnt = 0;
  Attempt(0);
}
void Solve() {
  int n = (int)s.length();
  lli res = 0;
  for (int i = 0; i < n; ++i) {
    string temp = "";
    int j;
    for (j = i; j < n; ++j) {
      temp.push_back(s[j]);
      if (Set.find(temp) == Set.end()) break;
    }
    res += n - j;
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  Generate();
  getline(cin, s);
  Solve();
}

### Prompt

Generate a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using lli = long long;
const int maxTrial = 10;
int cnt;
int x[maxTrial];
set<string> Set;
string s;
inline bool Check(int i) {
  for (int j = i - 2; j >= 0; j -= 2)
    if (x[i] == x[j] && x[i] == x[(i + j) / 2]) return false;
  return true;
}
void Print(int i) {
  string s;
  s.resize(i + 1);
  for (int j = 0; j <= i; ++j) s[j] = x[j] + '0';
  Set.insert(s);
}
void Attempt(int i) {
  for (x[i] = 0; x[i] <= 1; ++x[i])
    if (Check(i)) {
      Print(i);
      Attempt(i + 1);
    }
}
void Generate() {
  cnt = 0;
  Attempt(0);
}
void Solve() {
  int n = (int)s.length();
  lli res = 0;
  for (int i = 0; i < n; ++i) {
    string temp = "";
    int j;
    for (j = i; j < n; ++j) {
      temp.push_back(s[j]);
      if (Set.find(temp) == Set.end()) break;
    }
    res += n - j;
  }
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  Generate();
  getline(cin, s);
  Solve();
}

```