#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using ii = pair<int, int>;
using ll = pair<lint, lint>;
using ti = tuple<int, int, int>;
using il = pair<int, lint>;
const int N = 1e6 + 5;
const int M = 1e2 + 5;
int n, m, k;
char s[N];
int a[N], b[N];
bool ch(int l, int r) {
  for (int y = l; y <= r; y++) {
    int x = y - 1;
    int z = y + 1;
    while (l <= x && z <= r) {
      if (s[x] == s[y] && s[y] == s[z]) {
        return true;
      }
      --x;
      ++z;
    }
  }
  return false;
}
void input() {
  scanf("%s", s);
  n = strlen(s);
  lint ans = 0;
  for (int i = 0; i < n; i++) {
    bool o = false;
    int j;
    for (j = i; j < n; j++) {
      if (ch(i, j)) {
        o = true;
        break;
      }
    }
    if (o) {
      ans += (lint)n - j;
    }
  }
  printf("%lld\n", ans);
}
void go() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  go();
}

### Prompt

Please provide a cpp coded solution 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;
using lint = long long;
using ii = pair<int, int>;
using ll = pair<lint, lint>;
using ti = tuple<int, int, int>;
using il = pair<int, lint>;
const int N = 1e6 + 5;
const int M = 1e2 + 5;
int n, m, k;
char s[N];
int a[N], b[N];
bool ch(int l, int r) {
  for (int y = l; y <= r; y++) {
    int x = y - 1;
    int z = y + 1;
    while (l <= x && z <= r) {
      if (s[x] == s[y] && s[y] == s[z]) {
        return true;
      }
      --x;
      ++z;
    }
  }
  return false;
}
void input() {
  scanf("%s", s);
  n = strlen(s);
  lint ans = 0;
  for (int i = 0; i < n; i++) {
    bool o = false;
    int j;
    for (j = i; j < n; j++) {
      if (ch(i, j)) {
        o = true;
        break;
      }
    }
    if (o) {
      ans += (lint)n - j;
    }
  }
  printf("%lld\n", ans);
}
void go() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  go();
}

```