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 getArea(vector<int> &a, vector<int> &b) {
  long long x =
      *max_element(a.begin(), a.end()) - *min_element(a.begin(), a.end()) + 1;
  long long y =
      *max_element(b.begin(), b.end()) - *min_element(b.begin(), b.end()) + 1;
  return x * y;
}
long long Solve(vector<int> v, vector<int> a) {
  int cur = v[0], ind = 0;
  for (int i = 1; i < (v.size()); i++)
    if (v[i] > cur) {
      cur = v[i];
      ind = i;
    }
  for (int i = ind; i < (v.size()); i++) v[i]--;
  v.push_back(cur - 2);
  long long A = getArea(v, a);
  for (int i = ind; i < (v.size()); i++) v[i]++;
  v.pop_back();
  for (int i = 1; i < (v.size()); i++)
    if (v[i] < cur) {
      cur = v[i];
      ind = i;
    }
  for (int i = ind; i < (v.size()); i++) v[i]++;
  v.push_back(cur + 2);
  long long B = getArea(v, a);
  return min(A, B);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    vector<int> H, V;
    vector<long long> ans;
    H.push_back(0);
    V.push_back(0);
    int curH = 0, curV = 0;
    for (int i = 0; i < (s.size()); i++) {
      if (s[i] == 'A') {
        H.push_back(curH - 1);
        curH--;
      } else if (s[i] == 'D') {
        H.push_back(curH + 1);
        curH++;
      } else if (s[i] == 'S') {
        V.push_back(curV - 1);
        curV--;
      } else {
        V.push_back(curV + 1);
        curV++;
      }
    }
    ans.push_back(getArea(H, V));
    ans.push_back(Solve(H, V));
    ans.push_back(Solve(V, H));
    cout << *min_element(ans.begin(), ans.end()) << endl;
  }
}
