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 int T;
string s, a, b;
long long int d[200001];
long long int getX(char c) { return (c == 'D') - (c == 'A'); }
long long int getY(char c) { return (c == 'W') - (c == 'S'); }
long long int length(string a, char up) {
  long long int x, l;
  x = 0;
  l = 1;
  for (long long int i = 0; i < a.length(); ++i) {
    x += 2 * (a[i] == up) - 1;
    if (x < 0) {
      l -= x;
      x = 0;
    }
    l = max(x + 1, l);
  }
  return l;
}
long long int unidirectionalDelta(string a, char up) {
  d[0] = 0;
  long long int lastMax = 0;
  long long int firstMin = 0;
  bool single = true;
  for (long long int i = 1; i - 1 < a.length(); ++i) {
    single = i < a.length() ? a[i - 1] == a[i] : single;
    d[i] = d[i - 1] + (2 * (a[i - 1] == up) - 1);
    if (d[i] < d[firstMin]) {
      firstMin = i;
    }
    if (d[i] >= d[lastMax]) {
      lastMax = i;
    }
  }
  return (lastMax < firstMin) && (length(a, up) > 2);
}
long long int axisDelta(string a, char up, char down) {
  return unidirectionalDelta(a, up) || unidirectionalDelta(a, down);
}
int main() {
  cin >> T;
  for (long long int t = 0; t < T; ++t) {
    cin >> s;
    a = b = "";
    for (long long int i = 0; i < s.length(); ++i) {
      if (s[i] == 'W' or s[i] == 'S') {
        a += s[i];
      } else {
        b += s[i];
      }
    }
    long long int n = length(a, 'W');
    long long int m = length(b, 'D');
    long long int dn = axisDelta(a, 'W', 'S');
    long long int dm = axisDelta(b, 'A', 'D');
    long long int ans = min((n - dn) * m, n * (m - dm));
    cout << ans << endl;
  }
  return 0;
}
