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 ans;
int g, f, h, l, ppp[2][5], i;
int T, N, Sw[200010], Sh[200010];
char S[200010];
int main() {
  cin >> T;
  while (T--) {
    g = h = -192608170, f = l = 192608170;
    scanf("%s", S + 1), N = strlen(S + 1);
    for (i = 1; i <= N; ++i) {
      Sh[i] = Sh[i - 1], Sw[i] = Sw[i - 1];
      if (S[i] == 'W')
        Sh[i] = Sh[i - 1] + 1;
      else if (S[i] == 'S')
        Sh[i] = Sh[i - 1] - 1;
      else if (S[i] == 'D')
        Sw[i] = Sw[i - 1] + 1;
      else if (S[i] == 'A')
        Sw[i] = Sw[i - 1] - 1;
    }
    for (i = 0; i <= N; ++i) h = max(Sw[i], h), l = min(Sw[i], l);
    for (i = 0; i <= N; ++i) g = max(Sh[i], g), f = min(Sh[i], f);
    for (i = N; ~i; --i)
      if (Sh[i] == f) {
        ppp[0][4] = i;
        break;
      }
    for (i = N; ~i; --i)
      if (Sw[i] == l) {
        ppp[1][4] = i;
        break;
      }
    for (i = 0; i <= N; ++i)
      if (Sh[i] == g) {
        ppp[0][1] = i;
        break;
      }
    for (i = 0; i <= N; ++i)
      if (Sw[i] == h) {
        ppp[1][1] = i;
        break;
      }
    for (i = N; ~i; --i)
      if (Sh[i] == g) {
        ppp[0][3] = i;
        break;
      }
    for (i = N; ~i; --i)
      if (Sw[i] == h) {
        ppp[1][3] = i;
        break;
      }
    for (i = 0; i <= N; ++i)
      if (Sh[i] == f) {
        ppp[0][2] = i;
        break;
      }
    for (i = 0; i <= N; ++i)
      if (Sw[i] == l) {
        ppp[1][2] = i;
        break;
      }
    ans = 1ll * (l - h - 1) * (f - g - 1);
    if (ppp[0][3] < ppp[0][2] && Sh[ppp[0][3]] - Sh[ppp[0][2]] > 1)
      ans = min(ans, 1ll * (g - f) * (h - l + 1));
    if (ppp[1][3] < ppp[1][2] && Sw[ppp[1][3]] - Sw[ppp[1][2]] > 1)
      ans = min(ans, 1ll * (g - f + 1) * (h - l));
    if (ppp[0][4] < ppp[0][1] && Sh[ppp[0][4]] - Sh[ppp[0][1]] < -1)
      ans = min(ans, 1ll * (g - f) * (h - l + 1));
    if (ppp[1][4] < ppp[1][1] && Sw[ppp[1][4]] - Sw[ppp[1][1]] < -1)
      ans = min(ans, 1ll * (g - f + 1) * (h - l));
    cout << ans << endl;
    ppp[0][1] = ppp[1][1] = ppp[0][2] = ppp[1][2] = 192608170;
    ppp[0][3] = ppp[1][3] = ppp[0][4] = ppp[1][4] = -192608170;
  }
}
