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;
bool jeli(int mask, int i) { return (mask & (1 << i)) > 0; }
bool has(int mask, int n) {
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (jeli(mask, i) == jeli(mask, j)) {
        int k = j + j - i;
        if (k < n && jeli(mask, k) == jeli(mask, j)) return true;
      }
    }
  }
  return false;
}
bool fun(int n) {
  for (int i = 0; i < (1 << n); i++) {
    if (!has(i, n)) {
      for (int j = 0; j < n; j++) printf("%d", (i & (1 << j)) ? 1 : 0);
      printf("\n");
      return false;
    }
  }
  return true;
}
const int X = 10;
const int MAXN = 300100;
char s[MAXN];
int n;
void load() {
  scanf("%s", s);
  n = strlen(s);
}
bool ima(int a, int b) {
  for (int i = a; i <= b; i++) {
    for (int j = i + 1; j <= b; j++) {
      if (s[i] == s[j]) {
        int k = j + j - i;
        if (k <= b && s[k] == s[i]) {
          return true;
        }
      }
    }
  }
  return false;
}
void solve() {
  long long sol = 0;
  for (int i = 0; i < n; i++) {
    int j;
    for (j = i + 2; j < min(n, i + X); j++) {
      if (ima(i, j)) {
        sol++;
        break;
      }
    }
    if (j < n) sol += n - j - 1;
  }
  printf("%lld\n", sol);
}
int main() {
  load();
  solve();
  return 0;
}