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 N = 1e6 + 100;
int no_change(vector<int> v) {
  int mn = 1e9, mx = -1e9;
  for (int i = 0; i < v.size(); i++) {
    mn = min(mn, v[i]);
    mx = max(mx, v[i]);
  }
  return mx - mn + 1;
}
int can_change(vector<int> v) {
  int mn = 1e9, mx = -1e9;
  int mn_idx = -1, mx_idx = -1;
  for (int i = 0; i < v.size(); i++) {
    if (mn > v[i]) {
      mn = v[i];
      mn_idx = i;
    }
    if (mx < v[i]) {
      mx = v[i];
      mx_idx = i;
    }
  }
  int ans = no_change(v);
  for (int i = mn_idx; i < v.size(); i++) {
    v[i]++;
  }
  if (mn_idx == 0)
    v.insert(v.begin() + mn_idx, 0);
  else
    v.insert(v.begin() + mn_idx, v[mn_idx - 1] + 1);
  ans = min(ans, no_change(v));
  v.erase(v.begin() + mn_idx);
  for (int i = mn_idx; i < v.size(); i++) {
    v[i]--;
  }
  for (int i = mn_idx; i < v.size(); i++) {
    v[i]--;
  }
  if (mn_idx == 0)
    v.insert(v.begin() + mn_idx, 0);
  else
    v.insert(v.begin() + mn_idx, v[mn_idx - 1] - 1);
  ans = min(ans, no_change(v));
  v.erase(v.begin() + mn_idx);
  for (int i = mn_idx; i < v.size(); i++) {
    v[i]++;
  }
  for (int i = mx_idx; i < v.size(); i++) {
    v[i]++;
  }
  if (mx_idx == 0)
    v.insert(v.begin() + mx_idx, 0);
  else
    v.insert(v.begin() + mx_idx, v[mx_idx - 1] + 1);
  ans = min(ans, no_change(v));
  v.erase(v.begin() + mx_idx);
  for (int i = mx_idx; i < v.size(); i++) {
    v[i]--;
  }
  for (int i = mx_idx; i < v.size(); i++) {
    v[i]--;
  }
  if (mx_idx == 0)
    v.insert(v.begin() + mx_idx, 0);
  else
    v.insert(v.begin() + mx_idx, v[mx_idx - 1] - 1);
  ans = min(ans, no_change(v));
  v.erase(v.begin() + mx_idx);
  for (int i = mx_idx; i < v.size(); i++) {
    v[i]++;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    string s;
    cin >> s;
    vector<int> balance[2];
    balance[0].push_back(0);
    balance[1].push_back(0);
    int x = 0, y = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'W')
        x++;
      else if (s[i] == 'S')
        x--;
      else if (s[i] == 'A')
        y++;
      else
        y--;
      balance[0].push_back(x);
      balance[1].push_back(y);
    }
    long long ans1 = 1ll * can_change(balance[0]) * no_change(balance[1]);
    long long ans2 = 1ll * can_change(balance[1]) * no_change(balance[0]);
    cout << min(ans1, ans2) << "\n";
  }
  return 0;
}
