There are (n) rows of (m) people. Let the position in the (r)-th row and the (c)-th column be denoted by ((r, c)). Number each person starting from (1) in row-major order, i.e., the person numbered ((r-1)\cdot m+c) is initially at ((r,c)). The person at ((r, c)) decides to leave. To fill the gap, let the person who left be numbered (i). Each person numbered (j>i) will move to the position where the person numbered (j-1) is initially at. The following diagram illustrates the case where (n=2), (m=3), (r=1), and (c=2). Calculate the sum of the Manhattan distances of each person's movement. If a person was initially at ((r_0, c_0)) and then moved to ((r_1, c_1)), the Manhattan distance is (|r_0-r_1|+|c_0-c_1|). The first line contains a single integer (t) ((1\le t\le 10^4)) — the number of test cases. The only line of each testcase contains (4) integers (n), (m), (r), and (c) ((1\le r\le n\le 10^6), (1 \le c \le m \le 10^6)), where (n) is the number of rows, (m) is the number of columns, and ((r,c)) is the position where the person who left is initially at. For each test case, output a single integer denoting the sum of the Manhattan distances. For the first test case, the person numbered (2) leaves, and the distances of the movements of the person numbered (3), (4), (5), and (6) are (1), (3), (1), and (1), respectively. So the answer is (1+3+1+1=6). For the second test case, the person numbered (3) leaves, and the person numbered (4) moves. The answer is (1).