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;
tuple<int, int, int> findmax(vector<int> &arr) {
  int val = arr[0], left = 0, right = 0;
  for (size_t i = 0; i < arr.size(); i++) {
    if (arr[i] > val) {
      val = arr[i];
      left = right = i;
    } else if (arr[i] == val)
      right = i;
  }
  return make_tuple(val, left, right);
}
tuple<int, int, int> findmin(vector<int> &arr) {
  int val = arr[0], left = 0, right = 0;
  for (size_t i = 0; i < arr.size(); i++) {
    if (arr[i] < val) {
      val = arr[i];
      left = right = i;
    } else if (arr[i] == val)
      right = i;
  }
  return make_tuple(val, left, right);
}
int main() {
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    string s;
    vector<int> hor(1, 0), ver(1, 0);
    cin >> s;
    for (char ch : s) {
      if (ch == 'W')
        ver.push_back(ver.back() + 1);
      else if (ch == 'S')
        ver.push_back(ver.back() - 1);
      else if (ch == 'A')
        hor.push_back(hor.back() - 1);
      else
        hor.push_back(hor.back() + 1);
    }
    auto [hormax, hormax_left, hormax_right] = findmax(hor);
    auto [hormin, hormin_left, hormin_right] = findmin(hor);
    auto [vermax, vermax_left, vermax_right] = findmax(ver);
    auto [vermin, vermin_left, vermin_right] = findmin(ver);
    long long rls = (long long)(hormax - hormin + 1) * (vermax - vermin + 1);
    if (vermax_left > vermin_right || vermin_left > vermax_right)
      if (vermax - vermin + 1 > 2)
        rls = min(rls, (long long)(hormax - hormin + 1) * (vermax - vermin));
    if (hormax_left > hormin_right || hormin_left > hormax_right)
      if (hormax - hormin + 1 > 2)
        rls = min(rls, (long long)(hormax - hormin) * (vermax - vermin + 1));
    cout << rls << endl;
  }
  return 0;
}
