A. Squats

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本文介绍了一个有趣的编程问题：如何在最少的时间内调整站立和蹲下仓鼠的数量，使得两者数量相等。通过输入仓鼠初始状态，算法计算并输出所需的最小操作次数及调整后的状态。

Pasha has many hamsters and he makes them work out. Today, n hamsters (n is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.

For another exercise, Pasha needs exactly $\text{[math]}$ hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?

Input

The first line contains integer n (2 ≤ n ≤ 200; n is even). The next line contains n characters without spaces. These characters describe the hamsters' position: the i-th character equals 'X', if the i-th hamster in the row is standing, and 'x', if he is sitting.

Output

In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.

Sample test(s)

input
4
xxXx

output
1
XxXx

input
2
XX

output
1
xX

input
6
xXXxXx

output
0
xXXxXx
题意：X，x分别表示站着的和蹲着的仓鼠，想让站的的和蹲着的数量一样多，每操作一次需要1分钟，求最小操作数
#include<iostream>
using namespace std;
int main()
{
    int n;
    while(cin>>n)
    {
        string s;
        cin>>s;
        int num1=0;
        for(int i=0;i<n;i++)
        {
            if(s[i]=='X')
            num1++;
        }
        if(num1==n/2)
        {
            cout<<"0\n"<<s<<endl;
        }
        else if(num1>n/2)
        {
            cout<<num1-n/2<<endl;
            int i,k;
            for(i=0,k=0;i<n&&k<num1-n/2;i++)
            {
                if(s[i]=='X')
                {
                    s[i]='x';
                    k++;
                }
            }
            cout<<s<<endl;
        }
        else if(num1<n/2)
        {
            cout<<n/2-num1<<endl;
            int i,k;
            for(i=0,k=0;i<n&&k<n/2-num1;i++)
            {
                if(s[i]=='x')
                {
                    s[i]='X';
                    k++;
                }
            }
            cout<<s<<endl;
        }
    }
    return 0;
}