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 = 2e5 + 5;
int arr[N] = {};
int L[N][4] = {}, R[N][4] = {};
int now[4] = {};
void add(int x) {
  now[x] = max(0, now[x] - 1);
  now[x ^ 3]++;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int T;
  string str;
  cin >> T;
  while (T--) {
    cin >> str;
    int n = str.size();
    memset(L, 0, sizeof(L));
    memset(R, 0, sizeof(R));
    for (int i = 1; i <= n; i++) {
      if (str[i - 1] == 'W') arr[i] = 0;
      if (str[i - 1] == 'D') arr[i] = 1;
      if (str[i - 1] == 'S') arr[i] = 3;
      if (str[i - 1] == 'A') arr[i] = 2;
    }
    now[0] = now[1] = now[2] = now[3] = 0;
    for (int i = 1; i <= n; i++) {
      add(arr[i]);
      for (int j = 0; j < 4; j++) L[i][j] = now[j];
    }
    now[0] = now[1] = now[2] = now[3] = 0;
    for (int i = n; i > 0; i--) {
      add(arr[i] ^ 3);
      for (int j = 0; j < 4; j++) R[i][j] = now[j];
    }
    long long ans = 0x7f7f7f7f7f7f;
    for (int i = 1; i <= n + 1; i++) {
      for (int j = 0; j < 4; j++) {
        int t1 = L[i - 1][j], t2 = L[i - 1][j ^ 3];
        L[i - 1][j]++;
        L[i - 1][j ^ 3] = max(0, L[i - 1][j ^ 3] - 1);
        long long r = max(L[i - 1][0], R[i][0]) + max(L[i - 1][3], R[i][3]) + 1;
        long long c = max(L[i - 1][1], R[i][1]) + max(L[i - 1][2], R[i][2]) + 1;
        ans = min(ans, r * c);
        L[i - 1][j] = t1;
        L[i - 1][j ^ 3] = t2;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
