Codeforces Round #242 (Div. 2) A. Squats

A. Squats

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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

问最少多少步。使得X与x数量同样并输出

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<algorithm>
#define LL __int64
#define inf 0x3f3f3f3f
using namespace std;
char now[10];
char s[210];
bool bj;
int main()
{
    int n,m,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        bj=false;
        getchar();
        scanf("%s",s);
        k=strlen(s);
        int l=k;
        j=0;
        for(i=0; i<k; i++)
        {
            if(s[i]=='X')
                j++;
        }
        if(n/2==j)
        {
            printf("0\n");
            printf("%s\n",s);
        }
        else if(n/2<j)
        {
            printf("%d\n",j-n/2);
            int sum=0;
            for(i=0;i<l;i++)
            {
                if(sum==(j-n/2))
                break;
                if(s[i]=='X')
                {
                    s[i]='x';
                    sum++;
                }
            }
            printf("%s\n",s);
        }
        else
        {
            printf("%d\n",n/2-j);
            int sum=0;
            for(i=0;i<l;i++)
            {
                if(sum==(n/2-j))
                    break;
                if(s[i]=='x')
                {
                    s[i]='X';
                    sum++;
                }
            }
            printf("%s\n",s);
        }
    }
    return 0;
}