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;
long long h[200001];
long long v[200001];
pair<long long, long long> fh[200002];
pair<long long, long long> bh[200002];
pair<long long, long long> fv[200002];
pair<long long, long long> bv[200002];
void wk() {
  long long n;
  string s;
  long long ans = 10000000000007;
  long long i;
  cin >> s, n = s.length();
  for (i = 0; i <= n + 1; i++)
    fh[i] = bh[i] = fv[i] = bv[i] =
        pair<long long, long long>(10000000000007, -10000000000007);
  fh[0] = fv[0] = pair<long long, long long>(0, 0);
  long long cur = 0;
  for (i = 0; i < n; i++) {
    if (s[i] == 'A')
      cur--;
    else if (s[i] == 'D')
      cur++;
    h[i + 1] = cur;
  }
  cur = 0;
  for (i = 0; i < n; i++) {
    if (s[i] == 'W')
      cur--;
    else if (s[i] == 'S')
      cur++;
    v[i + 1] = cur;
  }
  for (i = 1; i <= n; i++) {
    fh[i].first = min(fh[i - 1].first, h[i]);
    fh[i].second = max(fh[i - 1].second, h[i]);
    fv[i].first = min(fv[i - 1].first, v[i]);
    fv[i].second = max(fv[i - 1].second, v[i]);
  }
  for (i = n; i >= 0; i--) {
    bh[i].first = min(bh[i + 1].first, h[i]);
    bh[i].second = max(bh[i + 1].second, h[i]);
    bv[i].first = min(bv[i + 1].first, v[i]);
    bv[i].second = max(bv[i + 1].second, v[i]);
  }
  long long cst = fv[n].second - fv[n].first + 1;
  for (i = 1; i <= n + 1; i++) {
    long long a = min(fh[i - 1].first, bh[i - 1].first - 1);
    long long b = max(fh[i - 1].second, bh[i - 1].second - 1);
    ans = min(ans, (b - a + 1) * cst);
    a = min(fh[i - 1].first, bh[i - 1].first + 1);
    b = max(fh[i - 1].second, bh[i - 1].second + 1);
    ans = min(ans, (b - a + 1) * cst);
  }
  cst = fh[n].second - fh[n].first + 1;
  for (i = 1; i <= n + 1; i++) {
    long long a = min(fv[i - 1].first, bv[i - 1].first - 1);
    long long b = max(fv[i - 1].second, bv[i - 1].second - 1);
    ans = min(ans, (b - a + 1) * cst);
    a = min(fv[i - 1].first, bv[i - 1].first + 1);
    b = max(fv[i - 1].second, bv[i - 1].second + 1);
    ans = min(ans, (b - a + 1) * cst);
  }
  cout << min(ans, (fv[n].second - fv[n].first + 1) *
                       (fh[n].second - fh[n].first + 1))
       << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  long long ii;
  cin >> t;
  for (ii = 1; ii <= t; ii++) wk();
  return 0;
}
