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;
pair<long long, long long> searchMinMax(vector<long long> &x) {
  long long pmin, pmax, vmin, vmax;
  vmin = 1e9;
  vmax = -1e9;
  pmin = pmax = -1;
  for (int i = 0, ThxDem = int(x.size()); i < ThxDem; ++i) {
    if (x[i] < vmin) {
      vmin = x[i];
      pmin = i;
    }
    if (x[i] >= vmax) {
      vmax = x[i];
      pmax = i;
    }
  }
  return make_pair(pmin, pmax);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, cad, cws, ansV, ansH, ans;
  string s;
  vector<long long> vws, vad;
  pair<long long, long long> pos, lad, lws;
  cin >> n;
  while (n--) {
    cin >> s;
    cad = cws = 0;
    lad = {0, 0};
    lws = {0, 0};
    vws.push_back(0);
    vad.push_back(0);
    for (int i = 0, ThxDem = int(s.size()); i < ThxDem; ++i) {
      if (s[i] == 'A') {
        cad += 1;
        vad.push_back(cad);
      }
      if (s[i] == 'D') {
        cad += -1;
        vad.push_back(cad);
      }
      if (s[i] == 'W') {
        cws += 1;
        vws.push_back(cws);
      }
      if (s[i] == 'S') {
        cws += -1;
        vws.push_back(cws);
      }
      lad = {min(lad.first, cad), max(lad.second, cad)};
      lws = {min(lws.first, cws), max(lws.second, cws)};
    }
    ansV = (lad.second - lad.first + 1);
    ansH = (lws.second - lws.first + 1);
    cerr << "ansV"
         << " = " << (ansV) << "\n";
    cerr << "ansH"
         << " = " << (ansH) << "\n";
    ans = ansV * ansH;
    pos = searchMinMax(vws);
    if (pos.first > pos.second) ans = min(ans, (max(ansH - 1, 2ll)) * (ansV));
    pos = searchMinMax(vad);
    if (pos.first > pos.second) ans = min(ans, (ansH) * (max(ansV - 1, 2ll)));
    for (int i = 0, ThxDem = int(vws.size()); i < ThxDem; ++i) vws[i] *= -1;
    for (int i = 0, ThxDem = int(vad.size()); i < ThxDem; ++i) vad[i] *= -1;
    pos = searchMinMax(vws);
    if (pos.first > pos.second) ans = min(ans, (max(2ll, ansH - 1)) * (ansV));
    pos = searchMinMax(vad);
    if (pos.first > pos.second) ans = min(ans, (ansH) * (max(2ll, ansV - 1)));
    cout << max(ans, 2ll) << "\n";
    vws.clear();
    vad.clear();
  }
  return 0;
}
