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;
const int INF = 1e9;
pair<int, int> solve(vector<int>& a) {
  int n = (int)a.size();
  int sum = 0, mx = 0, mn = 0;
  for (int i = 0; i < n; ++i) {
    sum += a[i];
    mx = max(mx, sum);
    mn = min(mn, sum);
  }
  int ans = mx - mn + 1;
  sum = 0;
  int lmx = -1, rmx, lmn = -1, rmn;
  if (sum == mx) {
    if (lmx == -1) lmx = 0;
    rmx = 0;
  }
  if (sum == mn) {
    if (lmn == -1) lmn = 0;
    rmn = 0;
  }
  for (int i = 0; i < n; ++i) {
    sum += a[i];
    if (sum == mx) {
      if (lmx == -1) lmx = i + 1;
      rmx = i + 1;
    }
    if (sum == mn) {
      if (lmn == -1) lmn = i + 1;
      rmn = i + 1;
    }
  }
  int delt = 0;
  if (rmn + 1 < lmx || rmx + 1 < lmn) delt = -1;
  return {ans, delt};
}
void solve() {
  string s;
  cin >> s;
  vector<int> a, b;
  for (char c : s) {
    if (c == 'W')
      a.push_back(1);
    else if (c == 'S')
      a.push_back(-1);
    else if (c == 'A')
      b.push_back(-1);
    else if (c == 'D')
      b.push_back(1);
  }
  pair<int, int> pa = solve(a), pb = solve(b);
  cout << min((pa.first + pa.second) * 1LL * pb.first,
              pa.first * 1LL * (pb.first + pb.second))
       << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve();
}
