Problem: George is a cat, so he really likes to play. Most of all he likes to play with his array of positive integers b. During the game, George modifies the array by using special changes. Let's mark George's current array as b1, b2, ..., b|b| (record |b| denotes the current length of the array). Then one change is a sequence of actions: 

  * Choose two distinct indexes i and j (1 ≤ i, j ≤ |b|; i ≠ j), such that bi ≥ bj. 
  * Get number v = concat(bi, bj), where concat(x, y) is a number obtained by adding number y to the end of the decimal record of number x. For example, concat(500, 10) = 50010, concat(2, 2) = 22. 
  * Add number v to the end of the array. The length of the array will increase by one. 
  * Remove from the array numbers with indexes i and j. The length of the array will decrease by two, and elements of the array will become re-numbered from 1 to current length of the array. 



George played for a long time with his array b and received from array b an array consisting of exactly one number p. Now George wants to know: what is the maximum number of elements array b could contain originally? Help him find this number. Note that originally the array could contain only positive integers.

Input

The first line of the input contains a single integer p (1 ≤ p < 10100000). It is guaranteed that number p doesn't contain any leading zeroes.

Output

Print an integer — the maximum number of elements array b could contain originally.

Examples

Input

9555


Output

4

Input

10000000005


Output

2

Input

800101


Output

3

Input

45


Output

1

Input

1000000000000001223300003342220044555


Output

17

Input

19992000


Output

1

Input

310200


Output

2

Note

Let's consider the test examples: 

  * Originally array b can be equal to {5, 9, 5, 5}. The sequence of George's changes could have been: {5, 9, 5, 5} → {5, 5, 95} → {95, 55} → {9555}. 
  * Originally array b could be equal to {1000000000, 5}. Please note that the array b cannot contain zeros. 
  * Originally array b could be equal to {800, 10, 1}. 
  * Originally array b could be equal to {45}. It cannot be equal to {4, 5}, because George can get only array {54} from this array in one operation. 



Note that the numbers can be very large.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long fexp(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % p;
    b /= 2;
    a = a * a % p;
  }
  return ans;
}
long long modmulti(long long a, long long b) {
  return ((a % 1000000007) * 1ll * (b % 1000000007)) % 1000000007;
}
long long modadd(long long a, long long b) {
  long long asdfgh =
      ((a % 1000000007) + (b % 1000000007) + 1000000007 + 1000000007) %
      1000000007;
  asdfgh = (asdfgh + 1000000007) % 1000000007;
  return asdfgh;
}
long long modpower(long long a, long long n) {
  if (n == 0) return 1;
  if (n == 1) return a % 1000000007;
  long long b = modpower(a, n / 2);
  b = modmulti(b, b);
  if (n % 2 == 0) return b;
  return modmulti(a, b);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long modinv(long long a) { return modpower(a, 1000000007 - 2); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void err() {
  cout << "-1\n";
  return;
}
template <typename ForwardIterator, typename T>
ForwardIterator first_less_than(ForwardIterator first, ForwardIterator last,
                                T value) {
  auto it = std::lower_bound(first, last, value);
  return (it == first ? last : --it);
}
const int N = 1e6 + 5;
struct vertex {
  long long parent;
  long long rank;
  long long comp;
  vector<long long> vec;
};
struct vertex ver[N];
long long a[N];
long long find(long long n) {
  long long x = n;
  while (x != ver[x].parent) {
    x = ver[x].parent;
  }
  long long y = n;
  long long z = ver[y].parent;
  while (y != z) {
    ver[y].parent = x;
    y = z;
    z = ver[z].parent;
  }
  return x;
}
void makeset(long long n) {
  ver[n].rank = 0;
  ver[n].parent = n;
  ver[n].comp = 1;
  ver[n].vec.push_back(n);
}
void initiate(long long n) {
  for (long long i = 0; i < n; i++) makeset(i);
}
void join(vector<long long>& x, vector<long long>& y) {
  x.insert(x.end(), y.begin(), y.end());
}
void uni(long long x, long long y) {
  long long px = find(x);
  long long py = find(y);
  if (px == py)
    return;
  else if (ver[px].rank > ver[py].rank) {
    ver[py].parent = px;
    ver[px].comp += ver[py].comp;
    join(ver[px].vec, ver[py].vec);
  } else if (ver[py].rank > ver[px].rank) {
    ver[px].parent = py;
    ver[py].comp += ver[px].comp;
    join(ver[py].vec, ver[px].vec);
  } else {
    ver[py].parent = px;
    ver[px].rank++;
    ver[px].comp += ver[py].comp;
    join(ver[px].vec, ver[py].vec);
  }
}
bool comp(string s1, string s2) {
  if (s1.size() > s2.size()) return 1;
  if (s2.size() > s1.size()) return 0;
  if (s1 >= s2) return 1;
  if (s2 > s1) return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    string p;
    cin >> p;
    string s1, s2;
    long long ans = 0;
    for (long long i = p.size() - 1; i >= 0; i--) {
      if (p[i] == '0') {
        s2.push_back(p[i]);
      } else {
        s2.push_back(p[i]);
        reverse(s2.begin(), s2.end());
        s1 = p.substr(0, i);
        if (comp(s1, s2)) {
          ans++;
          s2.clear();
        }
        reverse(s2.begin(), s2.end());
      }
    }
    cout << ans + 1;
  }
}