[plain] Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.

Input

Input file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow.

Output

For each test case, you should output its reverse number, one case per line.

Sample Input


3
12
-12
1200

Sample Output


21
-21
2100

Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:
1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.

Input

Input file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow.

Output

For each test case, you should output its reverse number, one case per line.

Sample Input

3
12
-12
1200

Sample Output

21
-21
2100
[plain]

#include  
  
 int main() 
 { 
     int i; 
     int j; 
     int n; 
     int tab;                        //标记末尾是否有零 
     int num; 
     int flag;                      //标记是否是正负数   
     int count; 
     int array[100];                 //存储各位的数值 
  
     scanf("%d", &n); 
     while(n--) 
     {   
         scanf("%d", &num); 
          
         j=0; 
         tab=1;                //假设末尾有零 
         flag=1;               // 正数为1 
         count=0; 
         if(num<0 ) 
         { 
             flag=0;            //负数 
             num=-1* num;          
         } 
  
         while(num) 
         { 
             if(num%10==0 && tab!=0)     
             { 
                count++;        //计算零的个数 
             } 
             else 
             { 
                array[j++]=num%10; 
                tab=0;      
             } 
             num/=10; 
         } 
  
         if(flag==0)      
         { 
             printf("-"); 
         } 
  
         for(i=0; i         { 
              printf("%d", array[i]); 
         } 
          
         for(i=0; i         { 
             printf("0"); 
         } 
         if(n>0)       
         { 
             printf("\n"); 
         } 
     } 
  
     return 0; 
 }