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;
using pii = pair<int, int>;
set<pii> A, B;
char b[201010];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    A.clear();
    B.clear();
    int x = 0, y = 0;
    scanf("%s", b);
    int n = strlen(b);
    A.insert({x, 0});
    B.insert({y, 0});
    for (int i = 0; i < n; i++) {
      if (b[i] == 'W')
        y++;
      else if (b[i] == 'S')
        y--;
      else if (b[i] == 'A')
        x++;
      else
        x--;
      A.insert({x, i + 1});
      B.insert({y, i + 1});
    }
    long long ans = 1LL * (prev(A.end())->first - A.begin()->first + 1) *
                    (prev(B.end())->first - B.begin()->first + 1);
    x = 0, y = 0;
    int mx = 0, Mx = 0, my = 0, My = 0;
    for (int i = 0; i < n; i++) {
      int a = A.begin()->first, b = prev(A.end())->first, c = B.begin()->first,
          d = prev(B.end())->first;
      ans = min(ans, 1LL * (max(Mx, b + 1) - min(mx, a + 1) + 1) *
                         (max(My, d) - min(my, c) + 1));
      ans = min(ans, 1LL * (max(Mx, b - 1) - min(mx, a - 1) + 1) *
                         (max(My, d) - min(my, c) + 1));
      ans = min(ans, 1LL * (max(Mx, b) - min(mx, a) + 1) *
                         (max(My, d + 1) - min(my, c + 1) + 1));
      ans = min(ans, 1LL * (max(Mx, b) - min(mx, a) + 1) *
                         (max(My, d - 1) - min(my, c - 1) + 1));
      A.erase({x, i});
      B.erase({y, i});
      if (::b[i] == 'W')
        y++;
      else if (::b[i] == 'S')
        y--;
      else if (::b[i] == 'A')
        x++;
      else
        x--;
      Mx = max(Mx, x);
      mx = min(mx, x);
      My = max(My, y);
      my = min(my, y);
    }
    printf("%lld\n", ans);
  }
}
