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;
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.length();
    int px[n + 1], py[n + 1];
    px[0] = py[0] = 0;
    for (int i = 0; i < n; i++) {
      px[i + 1] = px[i];
      py[i + 1] = py[i];
      if (s[i] == 'W') {
        py[i + 1] = py[i] + 1;
      } else if (s[i] == 'S') {
        py[i + 1] = py[i] - 1;
      } else if (s[i] == 'A') {
        px[i + 1] = px[i] + 1;
      } else {
        px[i + 1] = px[i] - 1;
      }
    }
    int ll[n + 1], rl[n + 1], ul[n + 1], dl[n + 1];
    ll[0] = rl[0] = ul[0] = dl[0] = 0;
    for (int i = 1; i <= n; i++) {
      ll[i] = min(ll[i - 1], px[i]);
      rl[i] = max(rl[i - 1], px[i]);
      ul[i] = max(ul[i - 1], py[i]);
      dl[i] = min(dl[i - 1], py[i]);
    }
    int lr[n + 1], rr[n + 1], ur[n + 1], dr[n + 1];
    lr[n] = rr[n] = px[n];
    ur[n] = dr[n] = py[n];
    for (int i = n - 1; i >= 0; i--) {
      lr[i] = min(lr[i + 1], px[i]);
      rr[i] = max(rr[i + 1], px[i]);
      ur[i] = max(ur[i + 1], py[i]);
      dr[i] = min(dr[i + 1], py[i]);
    }
    long long ans = 1ll * (rl[n] - ll[n] + 1) * (ul[n] - dl[n] + 1);
    for (int i = 0; i <= n; i++) {
      int l, r, u, d;
      l = min(ll[i], lr[i]);
      r = max(rl[i], rr[i]);
      u = max(ul[i], ur[i] + 1);
      d = min(dl[i], dr[i] + 1);
      ans = min(ans, 1ll * (r - l + 1) * (u - d + 1));
      u = max(ul[i], ur[i] - 1);
      d = min(dl[i], dr[i] - 1);
      ans = min(ans, 1ll * (r - l + 1) * (u - d + 1));
      l = min(ll[i], lr[i] - 1);
      r = max(rl[i], rr[i] - 1);
      u = max(ul[i], ur[i]);
      d = min(dl[i], dr[i]);
      ans = min(ans, 1ll * (r - l + 1) * (u - d + 1));
      l = min(ll[i], lr[i] + 1);
      r = max(rl[i], rr[i] + 1);
      ans = min(ans, 1ll * (r - l + 1) * (u - d + 1));
    }
    cout << ans << endl;
  }
  return 0;
}
