CF1787B.Number Factorization

传统题 时间 2000 ms 内存 256 MiB 5 尝试 1 已通过 1 标签

Number Factorization

Given an integer nn.

Consider all pairs of integer arrays aa and pp of the same length such that n=aipin = \prod a_i^{p_i} (i.e. a1p1a2p2a_1^{p_1}\cdot a_2^{p_2}\cdot\ldots) (ai>1;pi>0a_i \gt 1;p_i \gt 0) and aia_i is the product of some (possibly one) distinct prime numbers.

For example, for n=28=2271=4171n = 28 = 2^2\cdot 7^1 = 4^1 \cdot 7^1 the array pair a=[2,7]a = [2, 7], p=[2,1]p = [2, 1] is correct, but the pair of arrays a=[4,7]a = [4, 7], p=[1,1]p = [1, 1] is not, because 4=224=2^2 is a product of non-distinct prime numbers.

Your task is to find the maximum value of umaipium a_i \cdot p_i (i.e. a1p1+a2p2+a_1\cdot p_1 + a_2\cdot p_2 + \ldots) over all possible pairs of arrays aa and pp. Note that you do not need to minimize or maximize the length of the arrays.

Input

Each test contains multiple test cases. The first line contains an integer tt (1t10001 \le t \le 1000) — the number of test cases.

Each test case contains only one integer nn (2n1092 \le n \le 10^9).

Output

For each test case, print the maximum value of umaipium a_i \cdot p_i.

Note

In the first test case, 100=102100 = 10^2 so that a=[10]a = [10], p=[2]p = [2] when umaipium a_i \cdot p_i hits the maximum value 102=2010\cdot 2 = 20. Also, a=[100]a = [100], p=[1]p = [1] does not work since 100100 is not made of distinct prime factors.

In the second test case, we can consider 1010 as 10110^1, so a=[10]a = [10], p=[1]p = [1]. Notice that when 10=215110 = 2^1\cdot 5^1, umaipi=7um a_i \cdot p_i = 7.

Samples

7
100
10
864
130056192
1000000000
2
999999018
20
10
22
118
90
2
333333009

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