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 int MOD = 1e9 + 7;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto &x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
inline int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline int mul(int a, int b) { return (int)((long long)a * b % MOD); }
inline int binpow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b /= 2;
  }
  return res;
}
inline int inv(int a) { return binpow(a, MOD - 2); }
int gcd(int a, int b, int &x, int &y) {
  if (a == 0) {
    x = 0, y = 1;
    return b;
  }
  int x1, y1;
  int d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
const int N = 3e5 + 5;
int find(vector<int> &v, int x) {
  int id = lower_bound((v).begin(), (v).end(), x) - v.begin();
  for (int j = id + 1; j < v.size() && j <= id + 50; ++j) {
    int last = 2 * v[j] - v[id];
    auto it = lower_bound((v).begin(), (v).end(), last);
    if (it == v.end()) continue;
    int val = *it;
    if (val == last) return last;
  }
  return N;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  vector<int> pos[2];
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0')
      pos[0].push_back(i);
    else
      pos[1].push_back(i);
  }
  vector<int> mn(n + 1, n);
  long long ans = 0;
  for (int i = n - 1; i >= 0; --i) {
    mn[i] = mn[i + 1];
    if (s[i] == '0')
      mn[i] = min(mn[i], find(pos[0], i));
    else
      mn[i] = min(mn[i], find(pos[1], i));
    ans += n - mn[i];
  }
  cout << ans << "\n";
  return 0;
}