This time I need you to calculate the f(n) . (3<=n<=1000000)
f(n)= Gcd(3)+Gcd(4)+…+Gcd(i)+…+Gcd(n). Gcd(n)=gcd(C[n][1],C[n][2],……,C[n][n-1]) C[n][k] means the number of way to choose k things from n some things. gcd(a,b) means the greatest common divisor of a and b.
输入解释
There are several test case. For each test case:One integer n(3<=n<=1000000). The end of the in put file is EOF.
输出解释
For each test case: The output consists of one line with one integer f(n).