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>;
using vi = vector<int>;
using ll = long long;
const string WASD = "WASD";
const int D = 4, dx[D] = {1, 0, -1, 0}, dy[D] = {0, -1, 0, 1};
const int N = 2e5 + 2, INF = 1e9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string moves;
    cin >> moves;
    int n = int((moves).size()) + 1;
    vi x(n), y(n);
    for (int i = int(1); i < int(n); i++) {
      int dir = WASD.find(moves[i - 1]);
      x[i] = x[i - 1] + dx[dir];
      y[i] = y[i - 1] + dy[dir];
    }
    vi max_x_pref(n + 1), min_x_pref(n + 1);
    vi max_y_pref(n + 1), min_y_pref(n + 1);
    for (int i = int(0); i < int(n); i++) {
      max_x_pref[i + 1] = max(max_x_pref[i], x[i]);
      min_x_pref[i + 1] = min(min_x_pref[i], x[i]);
      max_y_pref[i + 1] = max(max_y_pref[i], y[i]);
      min_y_pref[i + 1] = min(min_y_pref[i], y[i]);
    }
    vi max_x_suf(n + 1, -INF), min_x_suf(n + 1, INF);
    vi max_y_suf(n + 1, -INF), min_y_suf(n + 1, INF);
    for (int i = int(n - 1); i >= int(0); i--) {
      max_x_suf[i] = max(max_x_suf[i + 1], x[i]);
      min_x_suf[i] = min(min_x_suf[i + 1], x[i]);
      max_y_suf[i] = max(max_y_suf[i + 1], y[i]);
      min_y_suf[i] = min(min_y_suf[i + 1], y[i]);
    }
    auto area = [](int x, int y) { return (ll)x * y; };
    ll mn_area = area(max_x_pref[n] - min_x_pref[n] + 1,
                      max_y_pref[n] - min_y_pref[n] + 1);
    for (int i = int(0); i < int(n - 1); i++)
      for (int d = int(0); d < int(D); d++) {
        int min_x = min_x_pref[i + 1], max_x = max_x_pref[i + 1];
        int min_y = min_y_pref[i + 1], max_y = max_y_pref[i + 1];
        min_x = min(min_x, min_x_suf[i] + dx[d]);
        max_x = max(max_x, max_x_suf[i] + dx[d]);
        min_y = min(min_y, min_y_suf[i] + dy[d]);
        max_y = max(max_y, max_y_suf[i] + dy[d]);
        mn_area = min(mn_area, area(max_x - min_x + 1, max_y - min_y + 1));
      }
    cout << mn_area << '\n';
  }
  return 0;
}
