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 long long MOD = 1000000007;
bool calc(int &dx, const string &s, char l, char r) {
  int minv = 0, maxv = 0, cur = 0;
  vector<int> traj;
  traj.push_back(0);
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == l) {
      cur--;
      minv = min(minv, cur);
    } else if (s[i] == r) {
      cur++;
      maxv = max(maxv, cur);
    }
    traj.push_back(cur);
  }
  dx = maxv - minv + 1;
  if (dx <= 2) return false;
  vector<int> occmin, occmax;
  for (int i = 0; i < traj.size(); i++) {
    int p = traj[i];
    if (p == maxv) occmax.push_back(i);
    if (p == minv) occmin.push_back(i);
  }
  if (maxv != 0 && occmin.back() < occmax[0]) return true;
  if (minv != 0 && occmax.back() < occmin[0]) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    int dx, dy;
    bool x_ok = calc(dx, s, 'A', 'D');
    bool y_ok = calc(dy, s, 'S', 'W');
    long long ans = (long long)dx * dy;
    if (x_ok) ans = min(ans, (long long)(dx - 1) * dy);
    if (y_ok) ans = min(ans, (long long)dx * (dy - 1));
    cout << ans << '\n';
  }
  return 0;
}
