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.util.*;
import java.lang.*;
import java.io.*;
 
public class GFG {
	public static void main (String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringBuffer sb = new StringBuffer();
		int Q = Integer.parseInt(br.readLine());
		for(int q = 0; q < Q; q++){
		    long area = 0;
		    String inp = br.readLine();
		    long[] lrdu = scanner(inp);
		    long d = lrdu[0];
		    long u = lrdu[1];
		    long l = lrdu[2];
		    long r = lrdu[3];
		    long hw = lrdu[4];
		    long hs = lrdu[5];
		    long hd = lrdu[6];
		    long ha = lrdu[7];
		    
		    long b1, b2, b3, b4;
		    b1 = b2 = b3 = b4 = 0;
		    if(hs != 0 || hw != 0) b1 = 1;
		    if(hw != hs) b2 = 1;
		    if(hd != 0 || ha != 0) b3 = 1;
		    if(hd != ha) b4 = 1;
		    
	        long y1 = Math.max(hw, hs);
            long y2 = Math.max(b1, y1 - b2);
            long x1 = Math.max(hd, ha);
            long x2 = Math.max(b3, x1 - b4);
            area = Math.min((y1 + 1) * (x2 + 1), (1 + y2) * (x1 + 1));
		    
		    sb.append(area + "\n");
		}    
		System.out.print(sb);
	}
	public static long[] scanner(String s){
	    long[] lrdu = new long[8];
	    long r, u;
	    r = u = 0;
	    for(int i = 0; i < s.length(); i++){
	        if(s.charAt(i) == 'W'){
	            u++;
	            if(u > lrdu[1]) lrdu[1] = u;
	        }
	        else if(s.charAt(i) == 'S'){
	            u--;
	            if(u < lrdu[0]) lrdu[0] = u;
	        }
	        else if(s.charAt(i) == 'D'){
	            r++;
	            if(r > lrdu[3]) lrdu[3] = r;
	        }
	        else{
	            r--;
	            if(r < lrdu[2]) lrdu[2] = r;
	        }
	        if(lrdu[4] < u - lrdu[0]) lrdu[4] = u - lrdu[0];
	        if(lrdu[5] < lrdu[1] - u) lrdu[5] = lrdu[1] - u;
	        if(lrdu[6] < r - lrdu[2]) lrdu[6] = r - lrdu[2];
	        if(lrdu[7] < lrdu[3] - r) lrdu[7] = lrdu[3] - r;
	    }
	    return lrdu;
	}
	public static long gcd(long a, long b){
	    if(b == 0) return a;
	    else return gcd(b, a % b);
	}
	public static long number_of_divisors(long n){
	    long divisors = 0;
	    for(long i = 1; i <= Math.sqrt(n); i++){
	        if((n % i == 0) && (n / i == i)) divisors += 1;
	        else if((n % i == 0) && (n / i != i)) divisors += 2;
	    }
	    return divisors;
	}
}

