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;
void solve() {
  string s;
  cin >> s;
  int n = s.length();
  vector<long long> x(n + 1, 0), y(n + 1, 0);
  for (int i = 0; i < n; i++) {
    x[i + 1] = x[i];
    y[i + 1] = y[i];
    if (s[i] == 'W')
      x[i + 1]++;
    else if (s[i] == 'A')
      y[i + 1]--;
    else if (s[i] == 'S')
      x[i + 1]--;
    else
      y[i + 1]++;
  }
  vector<long long> xlh(n + 1), xll(n + 1), ylh(n + 1), yll(n + 1);
  for (int i = 1; i <= n; i++) {
    xlh[i] = max(xlh[i - 1], x[i]);
    xll[i] = min(xll[i - 1], x[i]);
    ylh[i] = max(ylh[i - 1], y[i]);
    yll[i] = min(yll[i - 1], y[i]);
  }
  vector<long long> xrh(n + 1), xrl(n + 1), yrh(n + 1), yrl(n + 1);
  xrh[n] = xrl[n] = x[n];
  yrh[n] = yrl[n] = y[n];
  for (int i = n - 1; i >= 0; i--) {
    xrh[i] = max(xrh[i + 1], x[i]);
    xrl[i] = min(xrl[i + 1], x[i]);
    yrh[i] = max(yrh[i + 1], y[i]);
    yrl[i] = min(yrl[i + 1], y[i]);
  }
  long long ans = (xlh[n] - xll[n] + 1) * (ylh[n] - yll[n] + 1);
  for (int i = 0; i < n; i++) {
    ans = min(ans, (max({xlh[i], xrh[i] + 1, x[i] + 1}) -
                    min({xll[i], xrl[i] + 1, x[i] + 1}) + 1) *
                       (max(ylh[i], yrh[i]) - min(yll[i], yrl[i]) + 1));
    ans = min(ans, (max({xlh[i], xrh[i] - 1, x[i] - 1}) -
                    min({xll[i], xrl[i] - 1, x[i] - 1}) + 1) *
                       (max(ylh[i], yrh[i]) - min(yll[i], yrl[i]) + 1));
    ans = min(ans, (max(xlh[i], xrh[i]) - min(xll[i], xrl[i]) + 1) *
                       (max({ylh[i], yrh[i] - 1, y[i] - 1}) -
                        min({yll[i], yrl[i] - 1, y[i] - 1}) + 1));
    ans = min(ans, (max(xlh[i], xrh[i]) - min(xll[i], xrl[i]) + 1) *
                       (max({ylh[i], yrh[i] + 1, y[i] + 1}) -
                        min({yll[i], yrl[i] + 1, y[i] + 1}) + 1));
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}
