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 long long MOD = (long long)1e9 + 7;
struct node {
  int L;
  int R;
  int U;
  int D;
};
int t;
long long ans;
string s;
node ds[200001], de[200001];
pair<int, int> cur[200001], ways[4] = {make_pair(0, 1), make_pair(0, -1),
                                       make_pair(1, 0), make_pair(-1, 0)};
void gun(pair<int, int> coor, node& rec) {
  if (coor.first <= rec.L) rec.L = coor.first;
  if (coor.first >= rec.R) rec.R = coor.first;
  if (coor.second >= rec.U) rec.U = coor.second;
  if (coor.second <= rec.D) rec.D = coor.second;
}
void pun(pair<int, int> to, node& rec, pair<int, int> coor) {
  rec.L += to.first;
  rec.R += to.first;
  rec.U += to.second;
  rec.D += to.second;
  gun(coor, rec);
}
void fun(node& X, node& Y) {
  ans = min(ans, (max(X.R, Y.R) - min(X.L, Y.L) + 1) * 1LL *
                     (max(X.U, Y.U) - min(X.D, Y.D) + 1));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> s;
    int n = s.length();
    memset(ds, 0, sizeof(ds));
    cur[0].first = 0;
    cur[0].second = 0;
    for (int i = 1; i <= n; i++) {
      cur[i].first = cur[i - 1].first;
      cur[i].second = cur[i - 1].second;
      if (s[i - 1] == 'W') cur[i].second++;
      if (s[i - 1] == 'A') cur[i].first--;
      if (s[i - 1] == 'S') cur[i].second--;
      if (s[i - 1] == 'D') cur[i].first++;
      ds[i].L = ds[i - 1].L;
      ds[i].R = ds[i - 1].R;
      ds[i].U = ds[i - 1].U;
      ds[i].D = ds[i - 1].D;
      gun(cur[i], ds[i]);
    }
    de[n].L = cur[n].first;
    de[n].R = cur[n].first;
    de[n].U = cur[n].second;
    de[n].D = cur[n].second;
    for (int i = n - 1; i >= 0; i--) {
      de[i].L = de[i + 1].L;
      de[i].R = de[i + 1].R;
      de[i].U = de[i + 1].U;
      de[i].D = de[i + 1].D;
      gun(cur[i], de[i]);
    }
    ans = 1e18;
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j < 4; j++) {
        pair<int, int> tmp = ways[j];
        node tmp2 = de[i];
        pun(tmp, tmp2, cur[i]);
        fun(ds[i], tmp2);
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
