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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    vector<int> v, h;
    for (auto c : s) {
      if (c == 'W' or c == 'S')
        v.push_back(c == 'W' ? 1 : -1);
      else
        h.push_back(c == 'D' ? 1 : -1);
    }
    auto f = [](vector<int>& v, bool check = 0) {
      vector<int> p;
      if ((int)v.size() == 0) return 1;
      int pos = 0;
      p.push_back(0);
      for (auto x : v) {
        pos += x;
        p.push_back(pos);
      }
      if (p.size() == 0) return 1;
      int mx = p[0];
      int mn = p[0];
      mx = *max_element(p.begin(), p.end());
      mn = *min_element(p.begin(), p.end());
      if (!check) return mx - mn + 1;
      int ans = mx - mn + 1;
      auto premx = p;
      auto premn = p;
      auto sufmx = p;
      auto sufmn = p;
      for (int i = 1; i < (int)p.size(); ++i) {
        premx[i] = max(premx[i], premx[i - 1]);
        premn[i] = min(premn[i], premn[i - 1]);
      }
      for (int i = p.size() - 2; i >= 0; i--) {
        sufmx[i] = max(sufmx[i], sufmx[i + 1]);
        sufmn[i] = min(sufmn[i], sufmn[i + 1]);
      }
      for (int i = 0; i < (int)p.size(); ++i) {
        ans = min(
            ans, max(premx[i], sufmx[i] + 1) - min(premn[i], sufmn[i] + 1) + 1);
        ans = min(
            ans, max(premx[i], sufmx[i] - 1) - min(premn[i], sufmn[i] - 1) + 1);
      }
      return ans;
    };
    cout << min(1LL * f(v) * f(h, 1), 1LL * f(v, 1) * f(h)) << "\n";
  }
  return 0;
}
