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;
vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
void solve() {
  map<char, pair<int, int>> ma;
  ma['W'] = {0, 1};
  ma['A'] = {-1, 0};
  ma['S'] = {0, -1};
  ma['D'] = {1, 0};
  string s;
  cin >> s;
  int n = s.length();
  pair<int, int> pos[n + 1];
  pos[0] = {0, 0};
  for (int q = 0; q < n; q++) {
    pos[q + 1].first = pos[q].first + ma[s[q]].first;
    pos[q + 1].second = pos[q].second + ma[s[q]].second;
  }
  vector<vector<int>> extreme(n + 1);
  for (int q = 0; q < 2; q++) {
    extreme[n].push_back(pos[n].first);
    extreme[n].push_back(pos[n].second);
  }
  for (int q = n - 1; q >= 0; q--) {
    extreme[q].push_back(min(pos[q].first, extreme[q + 1][0]));
    extreme[q].push_back(min(pos[q].second, extreme[q + 1][1]));
    extreme[q].push_back(max(pos[q].first, extreme[q + 1][2]));
    extreme[q].push_back(max(pos[q].second, extreme[q + 1][3]));
  }
  long long ans = (long long)(extreme[0][2] - extreme[0][0] + 1) *
                  (extreme[0][3] - extreme[0][1] + 1);
  int lx = 0, ly = 0, hx = 0, hy = 0;
  for (int q = 1; q <= n; q++) {
    for (int w = 0; w < 4; w++) {
      int llx = min(lx, dir[w].first + extreme[q][0]);
      int lly = min(ly, dir[w].second + extreme[q][1]);
      int hhx = max(hx, dir[w].first + extreme[q][2]);
      int hhy = max(hy, dir[w].second + extreme[q][3]);
      llx = min(llx, pos[q - 1].first + dir[w].first);
      lly = min(lly, pos[q - 1].second + dir[w].second);
      hhx = max(hhx, pos[q - 1].first + dir[w].first);
      hhy = max(hhy, pos[q - 1].second + dir[w].second);
      ans = min(ans, (long long)(hhx - llx + 1) * (hhy - lly + 1));
    }
    lx = min(lx, pos[q].first);
    ly = min(ly, pos[q].second);
    hx = max(hx, pos[q].first);
    hy = max(hy, pos[q].second);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
