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;
const int maxn = 2e5 + 5;
const int INF = 1e9 + 9;
int T;
int n;
string s;
void ReadInput() {
  cin >> s;
  n = s.length();
}
void Solve() {
  int first = 0, second = 0;
  int l = 0, r = 0, up = 0, dw = 0;
  int fl = 0, fr = 0, fup = 0, fdw = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == 'W') {
      if (++second > up) {
        up = second;
        fdw = 0;
        fup = 1;
      } else if (second == up)
        fup = 1;
    }
    if (s[i] == 'A') {
      if (--first < l) {
        l = first;
        fr = 0;
        fl = 1;
      } else if (first == l)
        fl = 1;
    }
    if (s[i] == 'S') {
      if (--second < dw) {
        dw = second;
        fup = 0;
        fdw = 1;
      } else if (second == dw)
        fdw = 1;
    }
    if (s[i] == 'D') {
      if (++first > r) {
        r = first;
        fl = 0;
        fr = 1;
      } else if (first == r)
        fr = 1;
    }
  }
  first = r - l + 1;
  second = up - dw + 1;
  long long ans = 1LL * first * second, k = ans;
  if (first > 2 && fl * fr == 0) ans = k - second;
  if (second > 2 && fup * fdw == 0 && k - first < ans) ans = k - first;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> T;
  while (T-- > 0) {
    ReadInput();
    Solve();
  }
}
