Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items?

Input

The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100 characters. If the i-th character in the string is the letter "С", that means that the i-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the i-th character is the letter "P", than the i-th object on the wall is a photo.

Output

Print the only number — the minimum number of times Polycarpus has to visit the closet.

Examples

Input

CPCPCPC


Output

7


Input

CCCCCCPPPPPP


Output

4


Input

CCCCCCPPCPPPPPPPPPP


Output

6


Input

CCCCCCCCCC


Output

2

Note

In the first sample Polycarpus needs to take one item to the closet 7 times.

In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.

In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.

In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go).
#include <bits/stdc++.h>
using namespace std;
class gfg {
 public:
  unsigned int countSetBits(int n) {
    unsigned int count = 0;
    while (n) {
      n &= (n - 1);
      count++;
    }
    return count;
  }
} g;
vector<long long> prime;
long long int fact(int n);
int hexadecimalToDecimal(char hexVal[]);
vector<vector<long long int> > power(vector<vector<long long int> > &a,
                                     long long int n, long long int p);
long long int fact(long long int n, long long int p);
long long int power(long long int x, long long int n, long long int p);
long long int nCr(long long int n, long long int r, long long int p);
vector<vector<long long int> > multiply(vector<vector<long long int> > &a,
                                        vector<vector<long long int> > &b,
                                        long long int p);
void manipulated_seive(long long int N);
int gcd(int a, int b);
void generatePrimeFactors();
long long int calculateNoOFactors(long long int n);
void solve() {
  long long int answer;
  string n;
  cin >> n;
  vector<char> A;
  for (long long int i = 0; i < (n.size()); i++) {
    A.push_back((n[i]));
  }
  long long int ele = 1, countt = 0;
  for (long long int i = 1; i < A.size(); i++) {
    if (A[i] == A[i - 1] && ele < 5) {
      ele++;
    } else {
      ele = 1;
      countt++;
    }
  }
  cout << ++countt;
}
void template1() {
  long long int t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  cout << "\n";
  return 0;
}
long long int fact(int n) { return (n == 1 || n == 0) ? 1 : n * fact(n - 1); }
int hexadecimalToDecimal(char hexVal[]) {
  int len = strlen(hexVal);
  int base = 1;
  int dec_val = 0;
  for (int i = len - 1; i >= 0; i--) {
    if (hexVal[i] >= '0' && hexVal[i] <= '9') {
      dec_val += (hexVal[i] - 48) * base;
      base = base * 16;
    } else if (hexVal[i] >= 'A' && hexVal[i] <= 'F') {
      dec_val += (hexVal[i] - 55) * base;
      base = base * 16;
    }
  }
  return dec_val;
}
long long int fact(long long int n, long long int p) {
  long long int ans = 1;
  for (long long int i = 1; i <= n; i++) {
    ans *= i;
    ans %= p;
  }
  return ans;
}
long long int power(long long int x, long long int n, long long int p) {
  long long int res = 1;
  if (n == 1) return x % p;
  if (n % 2 == 1) res = x % p;
  long long int temp = power(x, n / 2, p);
  return res * ((temp * temp) % p) % p;
}
long long int inv(long long int a, long long int p) {
  return power(a, p - 2, p);
}
long long int nCr(long long int n, long long int r, long long int p) {
  if (r > n) return 0;
  long long int t1 = fact(n, p);
  long long int t2 = inv(fact(r, p), p);
  long long int t3 = inv(fact(n - r, p), p);
  return (((t1 * t2) % p) * t3) % p;
}
vector<vector<long long int> > multiply(vector<vector<long long int> > &a,
                                        vector<vector<long long int> > &b,
                                        long long int p) {
  int r1, c1, r2, c2;
  r1 = a.size();
  c1 = a[0].size();
  r2 = b.size();
  c2 = b[0].size();
  vector<vector<long long int> > result(r1, vector<long long int>(c2, 0));
  if (c1 != r2) return result;
  for (int i = 0; i < r1; i++) {
    for (int j = 0; j < c2; j++) {
      for (int k = 0; k < c1; k++) {
        result[i][j] += a[i][k] * b[k][j];
        result[i][j] %= p;
      }
    }
  }
  return result;
}
vector<vector<long long int> > power(vector<vector<long long int> > &a,
                                     long long int n, long long int p) {
  int r = a.size();
  vector<vector<long long int> > result(r, vector<long long int>(r, 0));
  for (int i = 0; i < r; i++) result[i][i] = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      result = multiply(result, a, p);
    }
    n /= 2;
    a = multiply(a, a, p);
  }
  return result;
}
const long long MAX_SIZE = 1000001;
vector<long long> isprime(MAX_SIZE, true);
vector<long long> SPF(MAX_SIZE);
void manipulated_seive(long long int N) {
  isprime[0] = isprime[1] = false;
  for (long long int i = 2; i < N; i++) {
    if (isprime[i]) {
      prime.push_back(i);
      SPF[i] = i;
    }
    for (long long int j = 0;
         j < (int)prime.size() && i * prime[j] < N && prime[j] <= SPF[i]; j++) {
      isprime[i * prime[j]] = false;
      SPF[i * prime[j]] = prime[j];
    }
  }
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
const int MAX = 1000001;
int factor[MAX] = {0};
void generatePrimeFactors() {
  factor[1] = 1;
  for (int i = 2; i < MAX; i++) factor[i] = i;
  for (int i = 4; i < MAX; i += 2) factor[i] = 2;
  for (int i = 3; i * i < MAX; i++) {
    if (factor[i] == i) {
      for (int j = i * i; j < MAX; j += i) {
        if (factor[j] == j) factor[j] = i;
      }
    }
  }
}
long long int calculateNoOFactors(long long int n) {
  if (n == 1) return 1;
  int ans = 1;
  int dup = factor[n];
  int c = 1;
  int j = n / factor[n];
  while (j != 1) {
    if (factor[j] == dup)
      c += 1;
    else {
      dup = factor[j];
      ans = ans * (c + 1);
      c = 1;
    }
    j = j / factor[j];
  }
  ans = ans * (c + 1);
  return ans;
}
