B'Recently Vova found n candy wrappers. He remembers that he bought x candies during the first day, 2x candies during the second day, 4x candies during the third day, ... , 2^{k-1} x candies during the k -th day. But there is an issue: Vova remembers neither x nor k but he is sure that x and k are positive integers and k > 1 . Vova will be satisfied if you tell him any positive integer x so there is an integer k>1 that x + 2x + 4x + ... + 2^{k-1} x = n . It is guaranteed that at least one solution exists. Note that k > 1 . You have to answer t independent test cases. The first line of the input contains one integer t ( 1 <= t <= 10^4 ) -- the number of test cases. Then t test cases follow. The only line of the test case contains one integer n ( 3 <= n <= 10^9 ) -- the number of candy wrappers Vova found. It is guaranteed that there is some positive integer x and integer k>1 that x + 2x + 4x + ... + 2^{k-1} x = n . Print one integer -- any positive integer value of x so there is an integer k>1 that x + 2x + 4x + ... + 2^{k-1} x = n . In the first test case of the example, one of the possible answers is x=1, k=2 . Then 1 cdot 1 + 2 cdot 1 equals n=3 . In the second test case of the example, one of the possible answers is x=2, k=2 . Then 1 cdot 2 + 2 cdot 2 equals n=6 . In the third test case of the example, one of the possible answers is x=1, k=3 . Then 1 cdot 1 + 2 cdot 1 + 4 cdot 1 equals n=7 . In the fourth test case of the example, one of the possible answers is x=7, k=2 . Then 1 cdot 7 + 2 cdot 7 equals n=21 . In the fifth test case of the example, one of the possible answers is x=4, k=3 . Then 1 cdot 4 + 2 cdot 4 + 4 cdot 4 equals n=28 . '...