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 ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    int x = 0, y = 0;
    int n = s.length();
    vector<pair<int, int>> kek(n + 1);
    kek[0] = {0, 0};
    for (int i = 0; i < n; i++) {
      char c = s[i];
      if (c == 'W') {
        x++;
      }
      if (c == 'S') {
        x--;
      }
      if (c == 'D') {
        y++;
      }
      if (c == 'A') {
        y--;
      }
      kek[i + 1] = {x, y};
    }
    vector<int> suf_x[2];
    vector<int> suf_y[2];
    suf_x[0].resize(n + 1);
    suf_x[1].resize(n + 1);
    suf_y[0].resize(n + 1);
    suf_y[1].resize(n + 1);
    suf_x[0][n] = suf_x[1][n] = kek[n].first;
    suf_y[0][n] = suf_y[1][n] = kek[n].second;
    for (int i = n - 1; i >= 0; i--) {
      suf_x[0][i] = min(suf_x[0][i + 1], kek[i].first);
      suf_x[1][i] = max(suf_x[1][i + 1], kek[i].first);
      suf_y[0][i] = min(suf_y[0][i + 1], kek[i].second);
      suf_y[1][i] = max(suf_y[1][i + 1], kek[i].second);
    }
    ll ans =
        ll(suf_x[1][0] - suf_x[0][0] + 1) * (suf_y[1][0] - suf_y[0][0] + 1);
    int lx = 0, rx = 0;
    int ly = 0, ry = 0;
    for (int i = 1; i <= n; i++) {
      ans = min(ans, ll(max(suf_x[1][i] + 1, max(rx, kek[i - 1].first + 1)) -
                        min(lx, suf_x[0][i] + 1) + 1) *
                         (max(suf_y[1][i], ry) - min(ly, suf_y[0][i]) + 1));
      ans =
          min(ans, ll(max(suf_x[1][i] - 1, rx) -
                      min(min(lx, kek[i - 1].first - 1), suf_x[0][i] - 1) + 1) *
                       (max(suf_y[1][i], ry) - min(ly, suf_y[0][i]) + 1));
      ans = min(ans,
                ll(max(suf_x[1][i], rx) - min(lx, suf_x[0][i]) + 1) *
                    (max(suf_y[1][i] - 1, ry) -
                     min(min(ly, kek[i - 1].second - 1), suf_y[0][i] - 1) + 1));
      ans = min(ans, ll(max(suf_x[1][i], rx) - min(lx, suf_x[0][i]) + 1) *
                         (max(suf_y[1][i] + 1, max(ry, kek[i - 1].second + 1)) -
                          min(ly, suf_y[0][i] + 1) + 1));
      lx = min(lx, kek[i].first);
      rx = max(rx, kek[i].first);
      ly = min(ly, kek[i].second);
      ry = max(ry, kek[i].second);
    }
    cout << ans << '\n';
  }
  return 0;
}
