Least Common Multiple

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Time limit: 1 Seconds Memory limit: 32768K
Total Submit: 2373 Accepted Submit: 906

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The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

Input

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

Output

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

Sample Input

2
3 5 7 15
6 4 10296 936 1287 792 1

Sample Output

105
10296

求最小公倍数.

Digital Roots

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Time limit: 1 Seconds Memory limit: 32768K
Total Submit: 8841 Accepted Submit: 1989

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Background

The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

Input

The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.

Output

For each integer in the input, output its digital root on a separate line of the output.

Example

Input

24
39
0
Output

6
3
这个题目比较简单,简单说一下:
给定一个数,求它的数根,即每一位的和组成一个新的数,直到所求的和是一个一位数为止.

求数根:例如39
是这样求到的39--->3+9=12--->1+2=3结束.
123456--->1+2+3+4+5+6=16=1+6=7结束.

scanf("%d",&n);
while(n>0)
{
scanf("%d",&num);
max=0;
for(i=0;i<num;i++)
scanf("%ld",&a[i]);

数还不够大.

感谢楼上的提出的建议.

1.这个不是竞赛贴,所以不存在奖励.即使说要有奖励,那怎么个奖励法,也不是我一个人说了算的.
2.的确,这些题目本身就是浙大ACM题.如果你们有自己的号,可以自己去做.
3.题目算是简单,但我也说过,这些不是竞赛题.再者,题目主要针对学C不是很久的朋友.一般有一个简单,一个稍微难点的.
如果出难的,我怕做的人也不是很多,这个就有违当时的初衷了.

谢谢.
希望大家对此有什么建议,我们都会考虑采纳.