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;
map<char, int> dx = {{'A', -1}, {'D', 1}};
map<char, int> dy = {{'W', 1}, {'S', -1}};
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  int t;
  cin >> t;
  for (int q = 0; q < t; q++) {
    string s;
    cin >> s;
    int n = (int)s.size();
    vector<int> x(n + 1);
    vector<int> y(n + 1);
    vector<int> minx(n + 1), miny(n + 1), maxx(n + 1), maxy(n + 1);
    for (int i = 0; i < n; i++) {
      x[i + 1] = x[i] + dx[s[i]];
      y[i + 1] = y[i] + dy[s[i]];
      minx[i + 1] = min(minx[i], x[i + 1]);
      maxx[i + 1] = max(maxx[i], x[i + 1]);
      miny[i + 1] = min(miny[i], y[i + 1]);
      maxy[i + 1] = max(maxy[i], y[i + 1]);
    }
    int sx = maxx[n] - minx[n] + 1;
    int sy = maxy[n] - miny[n] + 1;
    long long ans = (long long)sx * sy;
    int minx1 = x[n], maxx1 = x[n], miny1 = y[n], maxy1 = y[n];
    for (int i = n; i > 0; i--) {
      minx1 = min(minx1, x[i - 1]);
      maxx1 = max(maxx1, x[i - 1]);
      miny1 = min(miny1, y[i - 1]);
      maxy1 = max(maxy1, y[i - 1]);
      int cx = max(maxx[i - 1], maxx1 + 1) - min(minx[i - 1], minx1 + 1) + 1;
      ans = min(ans, (long long)cx * sy);
      cx = max(maxx[i - 1], maxx1 - 1) - min(minx[i - 1], minx1 - 1) + 1;
      ans = min(ans, (long long)cx * sy);
      int cy = max(maxy[i - 1], maxy1 + 1) - min(miny[i - 1], miny1 + 1) + 1;
      ans = min(ans, (long long)sx * cy);
      cy = max(maxy[i - 1], maxy1 - 1) - min(miny[i - 1], miny1 - 1) + 1;
      ans = min(ans, (long long)sx * cy);
    }
    cout << ans << '\n';
  }
}
