You have a string s — a sequence of commands for your toy robot. The robot is placed in some cell of a rectangular grid. He can perform four commands:

  * 'W' — move one cell up; 
  * 'S' — move one cell down; 
  * 'A' — move one cell left; 
  * 'D' — move one cell right. 



Let Grid(s) be the grid of minimum possible area such that there is a position in the grid where you can place the robot in such a way that it will not fall from the grid while running the sequence of commands s. For example, if s = DSAWWAW then Grid(s) is the 4 × 3 grid:

  1. you can place the robot in the cell (3, 2); 
  2. the robot performs the command 'D' and moves to (3, 3); 
  3. the robot performs the command 'S' and moves to (4, 3); 
  4. the robot performs the command 'A' and moves to (4, 2); 
  5. the robot performs the command 'W' and moves to (3, 2); 
  6. the robot performs the command 'W' and moves to (2, 2); 
  7. the robot performs the command 'A' and moves to (2, 1); 
  8. the robot performs the command 'W' and moves to (1, 1). 

<image>

You have 4 extra letters: one 'W', one 'A', one 'S', one 'D'. You'd like to insert at most one of these letters in any position of sequence s to minimize the area of Grid(s).

What is the minimum area of Grid(s) you can achieve?

Input

The first line contains one integer T (1 ≤ T ≤ 1000) — the number of queries.

Next T lines contain queries: one per line. This line contains single string s (1 ≤ |s| ≤ 2 ⋅ 10^5, s_i ∈ \{W, A, S, D\}) — the sequence of commands.

It's guaranteed that the total length of s over all queries doesn't exceed 2 ⋅ 10^5.

Output

Print T integers: one per query. For each query print the minimum area of Grid(s) you can achieve.

Example

Input


3
DSAWWAW
D
WA


Output


8
2
4

Note

In the first query you have to get string DSAWW\underline{D}AW.

In second and third queries you can not decrease the area of Grid(s).
import java.io.*;

public class WASDstring {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		for (int i = 0; i < t; i++) {
			String wasd = br.readLine();
			int maxV = 0, minV = 0, maxH = 0, minH = 0;
			int firstMaxV = 0, firstMinV = 0, lastMaxV = 0, lastMinV = 0, firstMaxH = 0, firstMinH = 0, lastMaxH = 0, lastMinH = 0;
			int currH = 0, currV = 0;
			for (int j = 0; j < wasd.length(); j++) {
				if (wasd.charAt(j) == 'W') {
					currV++;
					if (currV > maxV) {
						maxV = currV;
						firstMaxV = j;
						lastMaxV = j;
					}
					if (currV == maxV) {
						lastMaxV = j;
					}
				}
				if (wasd.charAt(j) == 'A') {
					currH--;
					if (currH < minH) {
						minH = currH;
						firstMinH = j;
						lastMinH = j;
					}
					if (currH == minH) {
						lastMinH = j;
					}
				}
				if (wasd.charAt(j) == 'S') {
					currV--;
					if (currV < minV) {
						minV = currV;
						firstMinV = j;
						lastMinV = j;
					}
					if (currV == minV) {
						lastMinV = j;
					}
				}
				if (wasd.charAt(j) == 'D') {
					currH++;
					if (currH > maxH) {
						maxH = currH;
						firstMaxH = j;
						lastMaxH = j;
					}
					if (currH == maxH) {
						lastMaxH = j;
					}
				}
			}
			long vertD = 1 + maxV - minV, horD = 1 + maxH - minH;
			long area = vertD * horD;
			if ((lastMinV < firstMaxV && (maxV > 1 || minV != 0))|| (lastMaxV < firstMinV && (minV < -1 || maxV != 0))) {
				area = (vertD - 1) * horD;
			}
			if ((lastMinH < firstMaxH && (maxH > 1 || minH != 0)) || (lastMaxH < firstMinH && (minH < -1 || maxH != 0))) {
				area = Long.min(area, vertD * (horD - 1));
			}
			System.out.println(area);
		}
	}
}
