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 uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &item) {
  out << '(' << item.first << ", " << item.second << ')';
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (const auto &item : v) out << item << ' ';
  return out;
}
pii get(string s) {
  if (s.empty()) return {1, 1};
  int xmin = 0, xmax = 0, x = 0, fm = 0, fM = 0, lm = 0, lM = 0;
  for (int i = 0; i < s.length(); ++i) {
    if (s[i] == 'W' || s[i] == 'A')
      ++x;
    else
      --x;
    if (x < xmin) xmin = x, fm = i + 1;
    if (x > xmax) xmax = x, fM = i + 1;
    if (x == xmin) lm = i + 1;
    if (x == xmax) lM = i + 1;
  }
  pii ret;
  ret.first = ret.second = xmax - xmin + 1;
  if (fm >= lM + 2 || fM >= lm + 2) --ret.second;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  string s;
  for (cin >> t; t; --t) {
    cin >> s;
    string ver, hor;
    for (auto c : s)
      if (c == 'W' || c == 'S')
        ver.push_back(c);
      else
        hor.push_back(c);
    auto r1 = get(ver);
    auto r2 = get(hor);
    cout << min(min(1LL * r1.first * r2.first, 1LL * r1.first * r2.second),
                1LL * r1.second * r2.first)
         << '\n';
  }
  return 0;
}
