The interval angle between the hour hand and the minute hand is always among 0 degree and 180 degree(including 0 and 180 degree). For example, when it's 12 o'clock, the angle of the two hands is 0 while 6:00 is 180 degree. Try to calculate any degree when it's between 12:00 to 11:59.

Input
The input data are of various situations. Every team consists of two numbers : the first number stands for the hour(great than 0 and less than or equal to 12) and second represents the minute( among [0, 59]). The input ends when the two numbers are both zeros.
Output
Print out the minimum angle between the two hands with the normal time form outputed.
Input Sample

12 0
12 30
6 0
3 0
0 0

Output Sample

At 12:00 the angle is 0.0 degrees.
At 12:30 the angle is 165.0 degrees.
At 6:00 the angle is 180.0 degrees.
At 3:00 the angle is 90.0 degrees.





#include <stdio.h>
#include <math.h>
int main()
{
    int h, m;
    float angle;
    while (scanf("%d %d", &h, &m) == 2)
    {
        if(h == 0 && m == 0)
        break;
        if (h == 12)
            h = 0;
        angle = fabs(m * 6 - h * 30 - m / 2);
        if (h == 0)
        h = 12;
        printf("At %d:%.2d the angle is %.1f degrees.\n", h, m, angle);
    }
    return 0;
}


输出都是对的，交上去就WA~~   无语啊~~

%.2d ==> %02d