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).
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
char s[N];
int n, r, c, mr, nr, mc, nc, tmr, tnr, tmc, tnc;
int main() {
  int Case;
  scanf("%d", &Case);
  while (Case--) {
    scanf("%s", s + 1), n = strlen(s + 1);
    r = c = mr = nr = mc = nc = tmr = tnr = tmc = tnc = 0;
    for (int i = 1; i <= n; ++i) {
      if (s[i] == 'W') ++r;
      if (s[i] == 'S') --r;
      if (s[i] == 'A') --c;
      if (s[i] == 'D') ++c;
      mr = max(mr, r), nr = min(nr, r), mc = max(mc, c), nc = min(nc, c);
      tmr = max(tmr, mr - r), tnr = max(tnr, r - nr);
      tmc = max(tmc, mc - c), tnc = max(tnc, c - nc);
    }
    int R1 = max(tmr, tnr), R2 = max((int)(tmr || tnr), R1 - (tmr != tnr));
    int C1 = max(tmc, tnc), C2 = max((int)(tmc || tnc), C1 - (tmc != tnc));
    ++R1, ++R2, ++C1, ++C2;
    printf("%lld\n", min(1LL * R1 * C2, 1LL * R2 * C1));
  }
  return 0;
}
