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 t;
int check(vector<int>& v) {
  int l = *min_element(v.begin(), v.end()),
      r = *max_element(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    if (v[i] == l) {
      l = i;
      break;
    }
  }
  for (int i = v.size() - 1; i >= 0; i--) {
    if (v[i] == r) {
      r = i;
      break;
    }
  }
  return l > r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    vector<int> X, Y;
    for (auto c : s) {
      if (c == 'W') Y.push_back(1);
      if (c == 'S') Y.push_back(-1);
      if (c == 'A') X.push_back(-1);
      if (c == 'D') X.push_back(1);
    }
    vector<int> px(X.size() + 1, 0), py(Y.size() + 1, 0), px2(X.size() + 1, 0),
        py2(Y.size() + 1, 0);
    for (int i = 1; i <= X.size(); i++)
      px[i] = px[i - 1] + X[i - 1], px2[i] = -px[i];
    for (int i = 1; i <= Y.size(); i++)
      py[i] = py[i - 1] + Y[i - 1], py2[i] = -py[i];
    int x = *max_element(px.begin(), px.end()) -
            *min_element(px.begin(), px.end()) + 1;
    int y = *max_element(py.begin(), py.end()) -
            *min_element(py.begin(), py.end()) + 1;
    long long res = 1ll * x * y;
    if (x > 2 && (check(px) || check(px2))) res = min(res, 1ll * (x - 1) * y);
    if (y > 2 && (check(py) || check(py2))) res = min(res, 1ll * x * (y - 1));
    cout << res << endl;
  }
}
