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 $c=2$.

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 $1$.