## ACM-ICPC 2018 南京赛区网络预赛 J题

A square-free integer is an integer which is indivisible by any square number except 11. For example, $6 = 2 \cdot 36=2⋅3$ is square-free, but $12 = 2^2 \cdot 312=22⋅3$ is not, because 2^222 is a square number. Some integers could be decomposed into product of two square-free integers, there may be more than one decomposition ways. For example, 6 = 1\cdot 6=6 \cdot 1=2\cdot 3=3\cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=aband n=ban=ba are considered different if a \not = ba̸=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aa and bb are square-free integers. The problem is calculating \sum_{i = 1}^nf(i)∑i=1nf(i).

### Input

The first line contains an integer T(T\le 20)T(T≤20), denoting the number of test cases.

For each test case, there first line has a integer n($n \le 2\cdot 10^7$)n(n≤2⋅107).

### Output

For each test case, print the answer $\sum_{i = 1}^n f(i)∑i=1nf(i)$.

### Hint

$$\sum_{i = 1}^8 f(i)=f(1)+ \cdots +f(8)∑i=18f(i)=f(1)+⋯+f(8) =1+2+2+1+2+4+2+0=14=1+2+2+1+2+4+2+0=14.$$

#### 题目来源

ACM-ICPC 2018 南京赛区网络预赛

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