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 form[200005];
int actions[2][200005], mem[2];
int brack_par[200005], no_change[2], changed[2];
int pre_min[200005], pre_max[200005];
void process(int par) {
  for (int i = 1; i <= mem[par]; i++) {
    brack_par[i] = brack_par[i - 1] + actions[par][i];
    pre_min[i] = min(pre_min[i - 1], brack_par[i]);
    pre_max[i] = max(pre_max[i - 1], brack_par[i]);
  }
  no_change[par] = changed[par] = pre_max[mem[par]] - pre_min[mem[par]] + 1;
  int nex_min = (int)1e9, nex_max = -(int)1e9;
  for (int i = mem[par]; i > 1; i--) {
    nex_min = min(nex_min, brack_par[i]);
    nex_max = max(nex_max, brack_par[i]);
    int all_max = max(nex_max + 1, pre_max[i - 1]);
    int all_min = min(nex_min + 1, pre_min[i - 1]);
    changed[par] = min(changed[par], all_max - all_min + 1);
    all_max = max(nex_max - 1, pre_max[i - 1]);
    all_min = min(nex_min - 1, pre_min[i - 1]);
    changed[par] = min(changed[par], all_max - all_min + 1);
  }
}
void solve() {
  scanf("%s", form + 1);
  int length = strlen(form + 1);
  mem[0] = mem[1] = 0;
  for (int i = 1; i <= length; i++)
    if (form[i] == 'W')
      actions[0][++mem[0]] = 1;
    else if (form[i] == 'S')
      actions[0][++mem[0]] = -1;
    else if (form[i] == 'D')
      actions[1][++mem[1]] = 1;
    else
      actions[1][++mem[1]] = -1;
  process(0), process(1);
  long long res = min((long long)changed[0] * no_change[1],
                      (long long)no_change[0] * changed[1]);
  printf("%lld\n", res);
}
int main() {
  int tester;
  scanf("%d", &tester);
  while (tester--) solve();
}
