Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33695 Accepted Submission(s): 15894

Problem Description

In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

Input

Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 10⁷ on each line.

Output

The output contains the number of digits in the factorial of the integers appearing in the input.

Sample Input

2 10 20

Sample Output

7 19

任意一个正整数a的位数等于(int)log10(a) + 1；

对于任意一个给定的正整数a，
  假设10^(x-1)<=a<10^x，那么显然a的位数为x位，
  又因为
  log10(10^(x-1))<=log10(a)<(log10(10^x))
  即x-1<=log10(a)<x
  则(int)log10(a)=x-1,
  即(int)log10(a)+1=x
  即a的位数是(int)log10(a)+1

那么我们要求的就是
(int)log10(A)+1，而：
	log10(A)
        =log10(1*2*3*......n)  （根据log10(a*b) = log10(a) + log10(b)有）
        =log10(1)+log10(2)+log10(3)+......+log10(n)

总结一下：n的阶乘的位数等于
		  (int)(log10(1)+log10(2)+log10(3)+......+log10(n)) + 1