Red was ejected. They were not the imposter.

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 $$$j \gt i$$$ 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 $$$c=2$$$.

$\text{[math]}$

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 $$$|r_0-r_1|+|c_0-c_1|$$$.

Input

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.

Output

For each test case, output a single integer denoting the sum of the Manhattan distances.

Note

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 $$$1$$$.

Sliding

Examples