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;
inline void showVec(vector<int> &vec) {
  for (int v : vec) cout << v << " ";
  cout << endl;
}
void solve() {
  string s;
  cin >> s;
  int n = s.length();
  vector<pair<int, int> > lrp(n + 2), udp(n + 2);
  vector<pair<int, int> > lrs(n + 2), uds(n + 2);
  vector<int> hd(n + 2), vd(n + 2);
  for (int i = 1; i <= n; ++i) {
    hd[i] = hd[i - 1], vd[i] = vd[i - 1];
    if (s[i - 1] == 'D')
      hd[i]++;
    else if (s[i - 1] == 'A')
      hd[i]--;
    else if (s[i - 1] == 'W')
      vd[i]++;
    else
      vd[i]--;
    lrp[i].first = min(lrp[i - 1].first, hd[i]);
    lrp[i].second = max(lrp[i - 1].second, hd[i]);
    udp[i].first = max(udp[i - 1].first, vd[i]);
    udp[i].second = min(udp[i - 1].second, vd[i]);
  }
  for (int i = n; i; --i) {
    lrs[i] = lrs[i + 1], uds[i] = uds[i + 1];
    if (s[i - 1] == 'D')
      lrs[i].first--, lrs[i].second++;
    else if (s[i - 1] == 'A')
      lrs[i].first++, lrs[i].second--;
    else if (s[i - 1] == 'W')
      uds[i].first++, uds[i].second--;
    else
      uds[i].first--, uds[i].second++;
    if (lrs[i].first < 0) lrs[i].first = 0;
    if (lrs[i].second < 0) lrs[i].second = 0;
    if (uds[i].first < 0) uds[i].first = 0;
    if (uds[i].second < 0) uds[i].second = 0;
  }
  int w = lrp[n].second - lrp[n].first + 1;
  for (int i = 1; i <= n; ++i) {
    int l = min(lrp[i].first, hd[i] + 1 - lrs[i + 1].first);
    int r = max(lrp[i].second, hd[i] + 1 + lrs[i + 1].second);
    w = min(w, r - l + 1);
  }
  for (int i = 1; i <= n; ++i) {
    int l = min(lrp[i].first, hd[i] - 1 - lrs[i + 1].first);
    int r = max(lrp[i].second, hd[i] - 1 + lrs[i + 1].second);
    w = min(w, r - l + 1);
  }
  int h = udp[n].first - udp[n].second + 1;
  for (int i = 1; i <= n; ++i) {
    int u = max(udp[i].first, vd[i] + 1 + uds[i + 1].first);
    int d = min(udp[i].second, vd[i] + 1 - uds[i + 1].second);
    h = min(h, u - d + 1);
  }
  for (int i = 1; i <= n; ++i) {
    int u = max(udp[i].first, vd[i] - 1 + uds[i + 1].first);
    int d = min(udp[i].second, vd[i] - 1 - uds[i + 1].second);
    h = min(h, u - d + 1);
  }
  cout << min(w * 1ll * (udp[n].first - udp[n].second + 1),
              (lrp[n].second - lrp[n].first + 1) * 1ll * h)
       << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
