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;
char a[200005];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf(" %s", a + 1);
    int len = strlen(a + 1);
    int now_x = 0, now_y = 0;
    int up = 0, down = 0, left = 0, right = 0;
    int t_up = 0, t_down = 0, t_left = 0, t_right = 0;
    for (int i = 1; i <= len; i++) {
      if (a[i] == 'D') now_y++;
      if (a[i] == 'W') now_x--;
      if (a[i] == 'A') now_y--;
      if (a[i] == 'S') now_x++;
      up = min(up, now_x);
      down = max(down, now_x);
      right = max(right, now_y);
      left = min(left, now_y);
      t_up = max(t_up, now_x - up);
      t_down = max(t_down, down - now_x);
      t_left = max(t_left, now_y - left);
      t_right = max(t_right, right - now_y);
    }
    int x = down - up + 1, y = right - left + 1;
    long long ans = 1ll * x * y;
    if (t_up != t_down) ans = min(ans, 1ll * max(x - 1, 2) * y);
    if (t_left != t_right) ans = min(ans, 1ll * x * max(y - 1, 2));
    printf("%lld\n", ans);
  }
  return 0;
}
