Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are n cockroaches living in Anatoly's room.

Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line toalternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.

Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of cockroaches.

The second line contains a string of length n, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.

Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.

In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.

In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.

In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0.

思路：

　　细想只有两种模式，一种brbrbr... 另一种rbrbrb... 只需要统计这两种模式下，需要的两种操作数中最小的一个，即是答案。

swap操作可以通过取Min来实现。接下来只需要统计每种模式下，需要变换的在字母个数就行（r2b or b2r）

Code：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <vector>

using namespace std;

const int Maxn = ;

char str[Maxn];

char str2[Maxn];

int main()

{

    int n;

    cin>>n;

    scanf("%s",str);

    int len = n, b2r = , r2b = , _b2r = , _r2b = ;

    for(int i = ; i < len; i ++){

        if(i %  == ){

            if(str[i] == 'r') r2b ++;

            if(str[i] == 'b') b2r ++;

        }else{

            if(str[i] == 'r') _r2b ++;

            if(str[i] == 'b') _b2r ++;

        }

    }

    cout<< min(max(r2b, _b2r), max(b2r, _r2b))<<endl;

    return ;

}