Problem:
Score : 300 points

Problem Statement

Two foxes Jiro and Saburo are playing a game called 1D Reversi. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa.

In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is B, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is W, it means that the color of the corresponding stone is white.

Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.

Constraints

1 ≦ |S| ≦ 10^5

Each character in S is B or W.

Input

The input is given from Standard Input in the following format:

S

Output

Print the minimum number of new stones that Jiro needs to place for his purpose.

Sample Input 1

BBBWW

Sample Output 1

1

By placing a new black stone to the right end of the row of stones, all white stones will become black. Also, by placing a new white stone to the left end of the row of stones, all black stones will become white.

In either way, Jiro's purpose can be achieved by placing one stone.

Sample Input 2

WWWWWW

Sample Output 2

0

If all stones are already of the same color, no new stone is necessary.

Sample Input 3

WBWBWBWBWB

Sample Output 3

9

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <cstdio>
#include <sstream>
#include <map>
#include <string>
#include <algorithm>
#include <queue>
#include <cmath>
#include <functional>
#include <set>
#include <ctime>
#include <random>
#include <chrono>
#include <cassert>
#include <tuple>
#include <utility>
using namespace std;

namespace {
  using Integer = long long; //__int128;
  template<class T, class S> istream& operator >> (istream& is, pair<T,S>& p){return is >> p.first >> p.second;}
  template<class T> istream& operator >> (istream& is, vector<T>& vec){for(T& val: vec) is >> val; return is;}
  template<class T> istream& operator ,  (istream& is, T& val){ return is >> val;}
  template<class T, class S> ostream& operator << (ostream& os, const pair<T,S>& p){return os << p.first << " " << p.second;}
  template<class T> ostream& operator << (ostream& os, const vector<T>& vec){for(size_t i=0; i<vec.size(); i++) os << vec[i] << (i==vec.size()-1?"":" "); return os;}
  template<class T> ostream& operator ,  (ostream& os, const T& val){ return os << " " << val;}

  template<class H> void print(const H& head){ cout << head; }
  template<class H, class ... T> void print(const H& head, const T& ... tail){ cout << head << " "; print(tail...); }
  template<class ... T> void println(const T& ... values){ print(values...); cout << endl; }

  template<class H> void eprint(const H& head){ cerr << head; }
  template<class H, class ... T> void eprint(const H& head, const T& ... tail){ cerr << head << " "; eprint(tail...); }
  template<class ... T> void eprintln(const T& ... values){ eprint(values...); cerr << endl; }

  class range{ Integer start_, end_, step_; public: struct range_iterator{ Integer val, step_; range_iterator(Integer v, Integer step) : val(v), step_(step) {} Integer operator * (){return val;} void operator ++ (){val += step_;} bool operator != (range_iterator& x){return step_ > 0 ? val < x.val : val > x.val;} }; range(Integer len) : start_(0), end_(len), step_(1) {} range(Integer start, Integer end) : start_(start), end_(end), step_(1) {} range(Integer start, Integer end, Integer step) : start_(start), end_(end), step_(step) {} range_iterator begin(){ return range_iterator(start_, step_); } range_iterator   end(){ return range_iterator(  end_, step_); } };

  inline string operator "" _s (const char* str, size_t size){ return move(string(str)); }
  constexpr Integer my_pow(Integer x, Integer k, Integer z=1){return k==0 ? z : k==1 ? z*x : (k&1) ? my_pow(x*x,k>>1,z*x) : my_pow(x*x,k>>1,z);}
  constexpr Integer my_pow_mod(Integer x, Integer k, Integer M, Integer z=1){return k==0 ? z%M : k==1 ? z*x%M : (k&1) ? my_pow_mod(x*x%M,k>>1,M,z*x%M) : my_pow_mod(x*x%M,k>>1,M,z);}
  constexpr unsigned long long operator "" _ten (unsigned long long value){ return my_pow(10,value); }

  inline int k_bit(Integer x, int k){return (x>>k)&1;} //0-indexed

  mt19937 mt(chrono::duration_cast<chrono::nanoseconds>(chrono::steady_clock::now().time_since_epoch()).count());

  template<class T> string join(const vector<T>& v, const string& sep){ stringstream ss; for(size_t i=0; i<v.size(); i++){ if(i>0) ss << sep; ss << v[i]; } return ss.str(); }

  inline string operator * (string s, int k){ string ret; while(k){ if(k&1) ret += s; s += s; k >>= 1; } return ret; }
}
constexpr long long mod = 9_ten + 7;

int main(){
  string s;
  cin >> s;

  s.erase( unique(s.begin(), s.end()), s.end());
  println(s.size()-1);

  return 0;
}