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 = 2e3 + 10;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
string s;
int maxl, maxr, minl, minr;
int solve(int c1, int c2) {
  maxl = maxr = minl = minr = 0;
  int pos = 0;
  int _max = 0, _min = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == c2) {
      pos++;
    } else if (s[i] == c1) {
      pos--;
    }
    if (pos > _max) {
      maxl = maxr = i;
      _max = pos;
    } else if (_max == pos) {
      maxr = i;
    }
    if (_min > pos) {
      _min = pos;
      minl = minr = i;
    } else if (_min == pos) {
      minr = i;
    }
  }
  return _max - _min + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> s;
    int len = solve('W', 'S');
    bool flag1 = false, flag2 = false;
    if ((maxr < minl || maxl > minr) && len > 2) {
      flag1 = true;
    }
    int len2 = solve('A', 'D');
    if ((maxr < minl || maxl > minr) && len2 > 2) {
      flag2 = true;
    }
    long long ans = (long long)len * len2;
    if (flag1) {
      ans = min((long long)(len - 1) * len2, ans);
    }
    if (flag2) {
      ans = min((long long)(len2 - 1) * len, ans);
    }
    cout << ans << endl;
  }
  return 0;
}
