#include <bits/stdc++.h>
using namespace std;
const int N = 300005;
int n, c, f[N], z, k;
char s[N];
long long ans;
vector<int> a[2];
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; ++i) {
    c = s[i] - '0';
    f[i] = -1;
    for (int j = a[c].size() - 1; j >= 0; --j) {
      z = a[c][j];
      k = i - z;
      if (s[z - k] == s[i]) {
        f[i] = z - k;
        break;
      }
      if (k + k >= i) break;
    }
    a[c].push_back(i);
  }
  int l = 0, r = 0;
  for (; l < n; ++l) {
    for (; r < n && (f[r] == -1 || f[r] < l); ++r)
      ;
    ans += r - l;
  }
  ans = 1ll * n * (n + 1) / 2 - ans;
  printf("%I64d", ans);
}

### Prompt

In CPP, your task is to solve 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;
const int N = 300005;
int n, c, f[N], z, k;
char s[N];
long long ans;
vector<int> a[2];
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; ++i) {
    c = s[i] - '0';
    f[i] = -1;
    for (int j = a[c].size() - 1; j >= 0; --j) {
      z = a[c][j];
      k = i - z;
      if (s[z - k] == s[i]) {
        f[i] = z - k;
        break;
      }
      if (k + k >= i) break;
    }
    a[c].push_back(i);
  }
  int l = 0, r = 0;
  for (; l < n; ++l) {
    for (; r < n && (f[r] == -1 || f[r] < l); ++r)
      ;
    ans += r - l;
  }
  ans = 1ll * n * (n + 1) / 2 - ans;
  printf("%I64d", ans);
}

```