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;
const long long LLMAX = 2e18;
const int MOD = 1e9 + 7;
const double eps = 1e-8;
const int MAXN = 1e6 + 10;
const int hmod1 = 0x48E2DCE7;
const int hmod2 = 0x60000005;
inline long long sqr(long long x) { return x * x; }
inline int sqr(int x) { return x * x; }
inline double sqr(double x) { return x * x; }
long long __gcd(long long a, long long b) {
  return b == 0 ? a : __gcd(b, a % b);
}
long long qpow(long long a, long long n) {
  long long sum = 1;
  while (n) {
    if (n & 1) sum = sum * a % MOD;
    a = a * a % MOD;
    n >>= 1;
  }
  return sum;
}
inline int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return (x > 0 ? 1 : -1);
}
int dis[MAXN];
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  long long n = s.size(), ans = 0;
  s = "+" + s;
  for (int i = 1; i <= n; i++) {
    for (int k = 1; k <= n && i + 2 * k <= n; k++)
      if (s[i] == s[i + k] && s[i + k] == s[i + 2 * k]) {
        dis[i] = i + 2 * k;
        break;
      }
  }
  for (int i = n; i >= 1; i--) {
    int p = dis[i];
    if (p == 0) p = n, dis[i] = n + 1;
    for (int j = i + 1; j <= p; j++) {
      if (dis[j] == n + 1) break;
      dis[i] = min(dis[i], dis[j]);
    }
  }
  for (int i = 1; i <= n; i++)
    if (dis[i] != n + 1) ans += (n - dis[i] + 1);
  cout << ans << endl;
  return 0;
}