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 long long INF = 1e18;
const long long inf = 1e9;
const long double eps = 1e-7;
const long double pi = acos(-1);
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
const int N = 1e7 + 1;
const int M = 1e3 + 11;
const int mod1 = 998244353;
const int mod2 = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int first = 0, second = 0, a = 0, b = 0, c = 0, d = 0;
    for (int i = 0; i < (int)s.size(); i++) {
      if (s[i] == 'W') second++, a = max(a, second), b = min(b, second);
      if (s[i] == 'S') second--, a = max(a, second), b = min(b, second);
      if (s[i] == 'A') first--, c = max(c, first), d = min(d, first);
      if (s[i] == 'D') first++, c = max(c, first), d = min(d, first);
    }
    first = 0, second = 0;
    long long ans = 1ll * (a - b + 1) * (c - d + 1);
    int d1 = 0, d2 = 0;
    for (int i = 0; i < (int)s.size(); i++) {
      if (s[i] == 'W') {
        if (second + 1 == a) second--, d1 = min(d1, second);
        second++;
        d2 = max(d2, second);
      }
      if (s[i] == 'S') second--, d1 = min(d1, second);
    }
    ans = min(ans, 1ll * (d2 - d1 + 1) * (c - d + 1));
    second = 0, d1 = 0, d2 = 0;
    for (int i = 0; i < (int)s.size(); i++) {
      if (s[i] == 'W') second++, d1 = max(d1, second);
      if (s[i] == 'S') {
        if (second - 1 == b) second++, d1 = max(d1, second);
        second--;
        d2 = min(d2, second);
      }
    }
    ans = min(ans, 1ll * (d1 - d2 + 1) * (c - d + 1));
    second = 0, d1 = 0, d2 = 0;
    for (int i = 0; i < (int)s.size(); i++) {
      if (s[i] == 'D') second++, d1 = max(d1, second);
      if (s[i] == 'A') {
        if (second - 1 == d) second++, d1 = max(d1, second);
        second--;
        d2 = min(d2, second);
      }
    }
    ans = min(ans, 1ll * (a - b + 1) * (d1 - d2 + 1));
    second = 0, d1 = 0, d2 = 0;
    for (int i = 0; i < (int)s.size(); i++) {
      if (s[i] == 'A') second--, d1 = min(d1, second);
      if (s[i] == 'D') {
        if (second + 1 == c) second--, d1 = min(d1, second);
        second++;
        d2 = max(d2, second);
      }
    }
    ans = min(ans, 1ll * (a - b + 1) * (d2 - d1 + 1));
    cout << ans << "\n";
  }
}
