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;
long long x[200010], y[200010];
long long a[200010], b[200010], c[200010], d[200010];
long long aa[200010], bb[200010], cc[200010], dd[200010];
void solve() {
  string s;
  cin >> s;
  if (s.size() == 1) {
    cout << 2 << endl;
    return;
  }
  x[0] = y[0] = 0;
  for (int i = 0; i < s.size(); i++) {
    x[i + 1] = x[i], y[i + 1] = y[i];
    if (s[i] == 'W') y[i + 1]--;
    if (s[i] == 'S') y[i + 1]++;
    if (s[i] == 'A') x[i + 1]--;
    if (s[i] == 'D') x[i + 1]++;
  }
  a[0] = b[0] = c[0] = d[0] = 0;
  for (int i = 1; i <= s.size(); i++) {
    a[i] = min(a[i - 1], x[i]);
    b[i] = max(b[i - 1], x[i]);
    c[i] = min(c[i - 1], y[i]);
    d[i] = max(d[i - 1], y[i]);
  }
  aa[s.size()] = bb[s.size()] = x[s.size()];
  cc[s.size()] = dd[s.size()] = y[s.size()];
  for (int i = s.size() - 1; i >= 0; i--) {
    aa[i] = min(aa[i + 1], x[i]);
    bb[i] = max(bb[i + 1], x[i]);
    cc[i] = min(cc[i + 1], y[i]);
    dd[i] = max(dd[i + 1], y[i]);
  }
  long long ans = 1e17;
  for (int i = 0; i + 1 <= s.size(); i++) {
    long long dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
    for (int r = 0; r < 4; r++) {
      long long xx = min(a[i], aa[i + 1] + dx[r]);
      xx = min(xx, x[i] + dx[r]);
      long long xxx = max(b[i], bb[i + 1] + dx[r]);
      xxx = max(xxx, x[i] + dx[r]);
      long long yy = min(c[i], cc[i + 1] + dy[r]);
      yy = min(yy, y[i] + dy[r]);
      long long yyy = max(d[i], dd[i + 1] + dy[r]);
      yyy = max(yyy, y[i] + dy[r]);
      ans = min(ans, (xxx - xx + 1) * (yyy - yy + 1));
    }
  }
  cout << ans << endl;
  return;
}
int main() {
  long long q;
  cin >> q;
  while (q--) solve();
  return 0;
}
