Consider Ø(n) as the Euler Totient Function for n. You will be given a positive integer N and you have to find the smallest positive integer n, n <= N for which the ratio n/Ø(n) is maximized.
First line of input consist of a single integer T denoting the total number of test case. Then T test cases follow. Each test case consists of a line with a positive integer N.
For each test case, in a new line print the smallest value of n, n <= N for which the ratio n/Ø(n) is maximized.
1 <= T <= 500
1 <= N <= 1012
Explanation: First test Case For n = 1, 2, 3, 4, 5 and 6 the values of the ratio are 1, 2, 1.5, 2, 1.25 and 3 respectively. The maximum is obtained at 6. Second test Case For n = 1 to 50, the maximum value of the ratio is 3.75 which is obtained at n = 30.