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;
int T;
int len;
vector<int> judge(char* ch, char ch1, char ch2) {
  int tp[200005];
  memset(tp, 0, sizeof(tp));
  int minn = 0, maxn = 0;
  tp[0] = 0;
  for (int i = 1; i <= len; i++) {
    tp[i] = tp[i - 1];
    if (ch[i - 1] == ch1) tp[i]++;
    if (ch[i - 1] == ch2) tp[i]--;
    minn = min(minn, tp[i]);
    maxn = max(maxn, tp[i]);
  }
  vector<int> ans;
  if (maxn == minn || maxn == (minn + 1))
    ans.push_back(0), ans.push_back(maxn - minn + 1);
  else {
    int pre = -1;
    int cnt = 0;
    for (int i = 0; i <= len; i++) {
      if (tp[i] == minn) {
        if (pre != 0) pre = 0, cnt++;
      } else if (tp[i] == maxn) {
        if (pre != 1) pre = 1, cnt++;
      }
    }
    if (cnt > 2)
      ans.push_back(0), ans.push_back(maxn - minn + 1);
    else
      ans.push_back(1), ans.push_back(maxn - minn + 1);
  }
  return ans;
}
int main() {
  while (~scanf("%d", &T)) {
    char ch[200005];
    while (T--) {
      scanf("%s", ch);
      len = strlen(ch);
      vector<int> tp1 = judge(ch, 'W', 'S');
      vector<int> tp2 = judge(ch, 'A', 'D');
      long long ans = 1ll * tp1[1] * tp2[1];
      if (tp1[0] == 1) ans = min(ans, 1ll * (tp1[1] - 1) * tp2[1]);
      if (tp2[0] == 1) ans = min(ans, 1ll * tp1[1] * (tp2[1] - 1));
      printf("%lld\n", ans);
    }
  }
  return 0;
}
