In the ruthless world of Blue Lock, Buratsuta 3 is a trio selected to overthrow the reigning champions and lead the Japan U-20 team to glory. Sae Itoshi has already secured his place as the first participant; the remaining two spots will be contested in the tough Side-B selection. To test the strategic abilities of the candidates, Buratsuta has posed the following challenge: You are given an array of (n) integers called "performance records" and (q) queries. Each query specifies a subarray (l, r). In this subarray, find all record values that occur strictly more than (\lfloor\tfrac{r - l + 1}{3}\rfloor) times. Each test consists of several test cases. The first line contains a single integer (t) ((1 \le t \le 10^4)) — the number of test cases. The following describes the test cases. The first line of each test case contains two integers (n) and (q) ((1 \le n, q \le 2\cdot10^5)) — the number of records and the number of queries. The second line of each test case contains (n) integers (a_1, a_2, \dots, a_n) ((1 \le a_i \le 10^9)) — the performance records. Each of the following (q) lines contains two integers (l) and (r) ((1 \le l \le r \le n)) — the boundaries of the query. It is guaranteed that the sum of (n) and sum of (q) over all test cases does not exceed (2 \cdot 10^5). For each query, output in one line all record values (in sorted order) that occur strictly more than (\lfloor\tfrac{r - l + 1}{3}\rfloor) times in the segment (l, r). If there are no such values, output (-1). In the second test case, the array is (a=1,1,2,3) and there are two queries: Query ((l,r)=(1,4)): The length of the segment (len=r-l+1=4), the threshold (\bigl\lfloor \frac{len}{3}\bigr\rfloor+1 = 2). Occurrences of numbers: (1\!\to\!2), (2\!\to\!1), (3\!\to\!1). Only the number (1) occurs at least (2) times, so the answer is: (1). Query $$$(l,r)=(2