Problem:
Score : 300 points

Problem Statement

N tiles are arranged in a row from left to right. The initial color of each tile is represented by a string S of length N.

The i-th tile from the left is painted black if the i-th character of S is 0, and painted white if that character is 1.

You want to repaint some of the tiles black or white, so that any two adjacent tiles have different colors.

At least how many tiles need to be repainted to satisfy the condition?

Constraints

1 \leq |S| \leq 10^5

S_i is 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum number of tiles that need to be repainted to satisfy the condition.

Sample Input 1

000

Sample Output 1

1

The condition can be satisfied by repainting the middle tile white.

Sample Input 2

10010010

Sample Output 2

3

Sample Input 3

0

Sample Output 3

0

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


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>

#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define ll_cin(X) scanf("%lld", &X)
#define ull_cin(X) scanf("%llu", &X)
#define lf_cin(X) scanf("%lf", &X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)

#define  ll_cout(X) printf("%lld\n",X)
#define  ull_cout(X) printf("%llu\n",X)



typedef long long  ll;
typedef unsigned long long  ull;

#define rept(x, s, n) for (int x = s; x < n; x++)

#define S_RTN(S)  s_cout(S);return 0


#define _itoa(A,N) sprintf(A, "%d", N);

int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);


int s_asorts(const void* a, const void* b);

//昇順
int s_asorts(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	for (i = 0; i < n; i++) {
		i_cin(*(a + i));
	}

	return 0;
}
int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}

long _max(long a, long b) {
	return a > b ? a : b;
}
long _min(long a, long b) {
	return a < b ? a : b;
}
int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

#pragma endregion



static char s[100001];
//coloring 124-C
int main(void) {
	
	int l, i, c = 0,c2=0;
	char w, b;

	scanf("%s",s); l = strlen(s);
	
	for (i = 0; i < l-1; i+=2) {
		if (s[i] != '0') c++;
		if (s[i + 1] != '1') c++;
		
		if (s[i] != '1') c2++;
		if (s[i+1] != '0') c2++;
	}
	if (i != l) {
	//	i_cout(i);
		if (s[i] != '0') c++;
		if (s[i] != '1') c2++;
	}
	

//	i_cout(c);
//	i_cout(c2);
	i_cout((int)_min(c,c2));
	return 0;
}



//static char s[500000];

//int main(void) {
//	int i,n, x, y, p1 = 200000,p2;
//	char q,tmp;
//	
//	s_cin(s[p1]);
//	p2 = p1+strlen(&s[p1])-1;
//	i_cin(n);
//
//
//	for (i = 0; i < n; i++) {
//		i_cin(x);
//		if (x == 1) { tmp = s[p1]; s[p2] = s[p1]; s[p2] = tmp; }
//		i_cin(y);scanf("%c",&q); scanf("%c", &q);
//		if (y == 1) { s[p1] = q; p1--;}
//		else { s[p2] = q; p2++; }
//	}
//	s_cout(&s[p1]);
//	return 0;
//}



// 93-C
//int main(void) {
//	int a, b, c;
//	int i,x, y, z, m, cnt = 0;
//	
//	i_cin3(a, b, c); 
//
//	x=abs(a-b);
//	y=abs(a-c);
//	z=abs(b-c);
//
//	m = _max(_max(x,y), z);
//	cnt = m / 2 + 1;
//	i_cout(cnt);
//	return 0;
//}



//#define _N 1000000000

//int main(void)
//{
//
//	int  n,i;
//	ull   tmp=1;
//
//	i_cin(n);
//
//	for (i = 1; i <= n; i++) {
//		if (!(tmp % ((ull)_N + 7)))
//			tmp = 1;
//		tmp *= (ull)i;
//	}
//	tmp %= ((ull)_N + 7);
//		
//	ull_cout(tmp);
//	return 0;
//}