Codeforces Round #373 (Div. 2)-B. Anatoly and Cockroaches

原题链接

B. Anatoly and Cockroaches

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

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.

Examples

input

5rbbrr

output

1

input

5bbbbb

output

2

input

3rbr

output

0

Note

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.

'r'和‘b'交替出现，就两种情况，第一个颜色要么是'r', 要么是'b', 算出变成这两种情况需要操作的次数，在求最小值

#include <bits/stdc++.h>#define maxn 200005#define MOD 1000000007using namespace std;typedef long long ll;char str[100005];int main(){//freopen("in.txt", "r", stdin);int n;scanf("%d%s", &n, str);int m = 'r' + 'b';int k1 = 'r', k2 = 'b';int r1 = 0, b1 = 0, r2 = 0, b2 = 0;for(int i = 0; str[i]; i++){if(k1 != str[i]){if(str[i] == 'r') r1++;else b1++;}if(k2 != str[i]){if(str[i] == 'r')  r2++;else b2++;}k1 = m - k1;k2 = m - k2;} cout << min(max(r1, b1), max(r2, b2)) << endl;return 0;}