How to Type


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5169    Accepted Submission(s): 2307


Problem Description


Pirates have finished developing the typing software. He called Cathy to test his typing software. She is good at thinking. After testing for several days, she finds that if she types a string by some ways, she will type the key at least. But she has a bad habit that if the caps lock is on, she must turn off it, after she finishes typing. Now she wants to know the smallest times of typing the key to finish typing a string.


 



Input


The first line is an integer t (t<=100), which is the number of test case in the input file. For each test case, there is only one string which consists of lowercase letter and upper case letter. The length of the string is at most 100.


 



Output


For each test case, you must output the smallest times of typing the key to finish typing this string.


 



Sample Input


3 Pirates HDUacm HDUACM


 



Sample Output


Hint

The string “Pirates”, can type this way, Shift, p, i, r, a, t, e, s, the answer is 8. The string “HDUacm”, can type this way, Caps lock, h, d, u, Caps lock, a, c, m, the answer is 8

The string "HDUACM", can type this way Caps lock h, d, u, a, c, m, Caps lock, the answer is 8

//简单dp题，可我做起来好艰难，哎哎，智商果然是不够用的。。。不过好像是第一道自己想出来的dp, 之前都是各种套模板，还是值得高兴的
//定义二维dp数组，第一维表示字符串长度，第二维表示大小写状态，长度为2. dp[i][0]代表i-1个字符且大写状态时最少敲击数，dp[i][1]代表i-1个字符且小写状态时最少敲击数
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <cctype>
using namespace std;

int main()
{
    int t;
    int dp[300][2];
    char s[300];

    scanf("%d", &t);
    while(t--)
    {
        scanf(" %s", s);

        int n = strlen(s);
        memset(dp, 0x3f, sizeof dp); //dp数组置为极大值
        dp[0][0] = 1,  dp[0][1] = 0; //初始状态赋值，因为打开大写，所以dp[0][0] = 1
        for(int i = 0; i < n; i++) //状态转移
        {
            if(isupper(s[i]))
            {
                dp[i+1][0] = min(dp[i][0] + 1, dp[i][1] + 2);
                dp[i+1][1] = min(dp[i][0] + 2, dp[i][1] + 2);
            }
            if(islower(s[i]))
            {
                dp[i+1][0] = min(dp[i][0] + 2, dp[i][1] + 2);
                dp[i+1][1] = min(dp[i][0] + 2, dp[i][1] + 1);
            }
        }

        printf("%d\n", min(dp[n][0] + 1, dp[n][1])); 
    }

    return 0;
}