Problem: There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`.

Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.

Constraints

* 1 \leq N \leq 2\times 10^5
* S is a string of length N consisting of `.` and `#`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the minimum number of stones that needs to be recolored.

Examples

Input

3
#.#


Output

1


Input

3
.#


Output

1


Input

5
.##.


Output

2


Input

9
.........


Output

0

Solution (in JAVA):

Solution:


import java.util.Scanner;

public class Main {

    public static void main(final String[] args) {
        try (final Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();
            String S = sc.next();
            // いったん白黒を数える、かつ境目の数も数える
            int border = 0;
            int b = 0;
            int w = 0;
            char prev = S.charAt(0);
            for (int i = 0; i < N; i++) {
                char c = S.charAt(i);
                b += c == '#' ? 1 : 0;
                w += c == '.' ? 1 : 0;
                if (c != prev) {
                    border += 1;
                }
                prev = c;
            }

            if (border == 1 && S.charAt(0) == '#') {
                // 境目が1つで黒からはじまる場合小さい方の数を採用
                System.out.println(Math.min(b, w));
            } else {
                // 境目ごとに左の黒、右の白の和を求めていく
                int leftB = 0;
                int rightW = w;
                int min = 0;
                prev = S.charAt(0);
                for (int i = 1; i < N; i++) {
                    char c = S.charAt(i);
                    leftB += prev == '#' ? 1 : 0;
                    rightW -= prev == '.' ? 1 : 0;
                    if (c != prev) {
                        int sum = leftB + rightW;
                        min = min == 0 ? sum : Math.min(min, sum);
                    }
                    prev = c;
                }
                System.out.println(min);
            }
        }
    }

}