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 ≤ 10

⁷ 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

 

123456=1.23456*10^5;
    log10(123456)=5.09151;
    log10(1.23456*10^5)=log10(1.23456)+log10(10^5)=0.09151+5;
    故int(log10(n))+1 就是n的位数 

 1、x的位数=(int)log10(x)+1；
 2、斯特林近似公式:n!≈sqrt(2*π*n)*(n/e)^n。

#include
      
       #include
       
        #include
        
         using namespace std;int main(){	int i,t,n;	double ans;	cin>>t;	while(t--){		cin>>n;		ans=0;		for(i=1;i<=n;i++) {			ans+=log10(double(i));		}		printf("%d\n",int(ans)+1);	}	return 0;}