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);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    auto get = [&](string t) {
      int x = 0, low = 0, high = 0;
      for (char &c : t) {
        if (c == 'W' || c == 'D') {
          x++;
        } else {
          x--;
        }
        low = min(low, x);
        high = max(high, x);
      }
      return (long long)high - low + 1;
    };
    string xs, ys;
    for (char &c : s) {
      if (c == 'W' || c == 'S') {
        xs += c;
      } else {
        ys += c;
      }
    }
    int len_x = get(xs), len_y = get(ys);
    long long ans = (long long)len_x * len_y;
    vector<int> pre_x(xs.size() + 1), pre_y(ys.size() + 1);
    for (int i = 0; i < (int)xs.size(); i++) {
      pre_x[i + 1] = pre_x[i] + (xs[i] == 'W' ? 1 : -1);
    }
    for (int i = 0; i < (int)ys.size(); i++) {
      pre_y[i + 1] = pre_y[i] + (ys[i] == 'D' ? 1 : -1);
    }
    int mi = min_element(pre_x.begin(), pre_x.end()) - pre_x.begin();
    int ma = max_element(pre_x.begin(), pre_x.end()) - pre_x.begin();
    if (mi > 0) {
      ans =
          min(ans, get(xs.substr(0, mi - 1) + 'W' + xs.substr(mi - 1)) * len_y);
    }
    if (ma > 0) {
      ans =
          min(ans, get(xs.substr(0, ma - 1) + 'S' + xs.substr(ma - 1)) * len_y);
    }
    mi = min_element(pre_y.begin(), pre_y.end()) - pre_y.begin();
    ma = max_element(pre_y.begin(), pre_y.end()) - pre_y.begin();
    if (mi > 0) {
      ans =
          min(ans, len_x * get(ys.substr(0, mi - 1) + 'D' + ys.substr(mi - 1)));
    }
    if (ma > 0) {
      ans =
          min(ans, len_x * get(ys.substr(0, ma - 1) + 'A' + ys.substr(ma - 1)));
    }
    cout << ans << '\n';
  }
  return 0;
}
