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 mxn = 2005;
const long long mxm = 1;
const long long INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const double eps = 1e-7;
const double pii = 3.1415926535898;
long long gcd(long long x, long long y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ans;
}
long long lowbit(long long x) { return (x & (-x)); }
long long qmul(long long a, long long b) {
  long long ans = 0;
  while (b) {
    if (b & 1) ans = (ans + a) % mod;
    b >>= 1;
    a = (a + a) % mod;
  }
  return ans;
}
string s;
long long le[300005], r[300005];
signed main(void) {
  cin >> s;
  long long l = s.size();
  long long ans = 0;
  for (long long i = 0; i < 300005; i++) le[i] = 3000005, r[i] = -1;
  for (long long i = 0; i < l; i++) {
    for (long long k = 1; i + 2 * k < l; k++) {
      if (s[i] == s[i + k] && s[i + k] == s[i + 2 * k]) {
        le[i + 1] = min(le[i + 1], i + 2 * k);
        break;
      }
    }
  }
  for (long long i = l; i >= 1; i--) {
    le[i] = min(le[i], le[i + 1]);
  }
  for (long long i = 1; i <= l; i++) {
    if (le[i] != 3000005) ans = ans + (l - le[i]);
  }
  printf("%lld\n", ans);
}