For a permutation(^{\text{∗}}) (r) of length (m), we call an index (i) ((1\le i\le m)) ugly if and only if (i=\max(\{r_1,r_2,\ldots,r_i\})). Simons has a permutation (p) of length (n), and he can perform the following operation on (p) at most once : Choose two indices (i) and (j) ((1\le i\ne j\le n)), then swap (p_i) and (p_j). Find a permutation (q) that can be obtained from (p) by performing the above operation at most once, such that the number of ugly indices in (q) is minimized. (^{\text{∗}})A permutation of length (n) is an array consisting of (n) distinct integers from (1) to (n) in arbitrary order. For example, (2,3,1,5,4) is a permutation, but (1,2,2) is not a permutation ((2) appears twice in the array), and (1,3,4) is also not a permutation ((n=3) but there is (4) in the array). Each test contains multiple test cases. The first line contains the number of test cases (t) ((1 \le t \le 100)). The description of the test cases follows. The first line contains an integer (n) ((1\le n\le 500)) — the length of (p). The second line contains (n) integers (p_1,p_2,\ldots,p_n) ((1\le p_i\le n), all (p_i)-s are distinct) — the elements of (p). For each test case, print (n) integers (q_1,q_2,\ldots,q_n) — the permutation you found. If there are multiple possible permutations, you may output any. In the first test case, Simons can obtain only two possible permutations: (1,2) and (2,1). For permutation (1,2), we can see (1=\max(\{1\})) and (2=\max(\{1,2\})). So there are two ugly indices. For permutation (2,1), we can see (1\ne\max(\{2\})) and (2=\max(\{2,1\})). So there is only one ugly index. Thus, permutation (2,1) has the minimum count of ugly indices. In the second test case, Simons can obtain permutation (2,4,1,3) by choosing indices $$$2$