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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    string s;
    cin >> s;
    vector<int> a;
    auto Solve = [&]() {
      int n = a.size();
      if (n == 0) return pair<long long, long long>{1, 1};
      if (n == 1) return pair<long long, long long>{2, 2};
      long long left = 0, right = 0, cur = 0;
      int mxi = 0, mni = 0;
      for (int i = 0; i < n; ++i) {
        cur += a[i];
        if (cur < left) {
          left = cur;
          mni = i;
        }
        if (cur > right) {
          right = cur;
          mxi = i;
        }
      }
      long long ans = right - left + 1;
      pair<long long, long long> out;
      out.first = ans;
      cur = 0;
      left = 0;
      right = 0;
      for (int i = 0; i < n; ++i) {
        if (i != mxi) {
          cur += a[i];
        }
        left = min(left, cur);
        right = max(right, cur);
      }
      ans = min(ans, right - left + 1);
      cur = 0;
      left = 0;
      right = 0;
      for (int i = 0; i < n; ++i) {
        if (i != mni) {
          cur += a[i];
        }
        left = min(left, cur);
        right = max(right, cur);
      }
      ans = min(ans, right - left + 1);
      out.second = ans;
      return out;
    };
    for (int i = 0; i < (int)s.size(); ++i) {
      if (s[i] == 'W') {
        a.push_back(1);
      } else if (s[i] == 'S') {
        a.push_back(-1);
      }
    }
    auto height = Solve();
    a.clear();
    for (int i = 0; i < (int)s.size(); ++i) {
      if (s[i] == 'A') {
        a.push_back(1);
      } else if (s[i] == 'D') {
        a.push_back(-1);
      }
    }
    auto width = Solve();
    long long ans = height.first * width.first;
    ans = min(ans, height.first * width.second);
    ans = min(ans, height.second * width.first);
    cout << ans << "\n";
  }
  return 0;
}
