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;
using ll = long long;
const int inf = 1e9 + 7;
const ll longinf = 1LL << 60;
const ll mod = 1e9 + 7;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    vector<ll> yoko(1, 0), tate(1, 0);
    int ct = 0, cy = 0;
    for (int i = (int)(0); i < (int)(n); ++i) {
      if (s[i] == 'W') tate.push_back(++ct);
      if (s[i] == 'S') tate.push_back(--ct);
      if (s[i] == 'A') yoko.push_back(++cy);
      if (s[i] == 'D') yoko.push_back(--cy);
    }
    ll ans[2][2];
    for (int k = (int)(0); k < (int)(2); ++k) {
      int m = yoko.size();
      vector<ll> fa(m + 1, -inf), fi(m + 1, inf), ba(m + 1, -inf),
          bi(m + 1, inf);
      for (int i = (int)(0); i < (int)(m); ++i) {
        fa[i + 1] = max(fa[i], yoko[i]);
        fi[i + 1] = min(fi[i], yoko[i]);
        ba[m - i - 1] = max(ba[m - i], yoko[m - i - 1]);
        bi[m - i - 1] = min(bi[m - i], yoko[m - i - 1]);
      }
      ans[k][0] = fa[m] - fi[m] + 1;
      ans[k][1] = inf;
      for (int i = (int)(0); i < (int)(m); ++i) {
        ans[k][1] = min(ans[k][1],
                        max({yoko[i] + 1, fa[i + 1], ba[i + 1] + 1}) -
                            min({yoko[i] + 1, fi[i + 1], bi[i + 1] + 1}) + 1);
        ans[k][1] = min(ans[k][1],
                        max({yoko[i] + 1, fa[i + 1], ba[i + 1] - 1}) -
                            min({yoko[i] + 1, fi[i + 1], bi[i + 1] - 1}) + 1);
      }
      yoko.swap(tate);
    }
    cout << min({ans[0][0] * ans[1][0], ans[0][1] * ans[1][0],
                 ans[0][0] * ans[1][1]})
         << "\n";
  }
  return 0;
}
