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;
long long l[200002], c[200002];
long long minx, maxx, miny, maxy;
string s;
bool check(long long l[], int n) {
  int Min = 0, Max = 0;
  int last = 0, first = 0;
  for (int i = 1; i <= n; i++) {
    if (l[i] < Min) {
      first = i;
      Min = l[i];
    }
    if (l[i] >= Max) {
      last = i;
      Max = l[i];
    }
  }
  if (last < first) return 1;
  Min = 0, Max = 0;
  first = last = 0;
  for (int i = 1; i <= n; i++) {
    if (l[i] > Max) {
      first = i;
      Max = l[i];
    }
    if (l[i] <= Min) {
      Min = l[i];
      last = i;
    }
  }
  if (last < first) return 1;
  return 0;
}
int main() {
  cin >> t;
  while (t--) {
    cin >> s;
    minx = maxx = miny = maxy = 0;
    long long x = 0, y = 0;
    for (int i = 0; i < s.size(); i++) {
      if (s[i] == 'D') c[i + 1] = c[i] + 1;
      if (s[i] == 'A') c[i + 1] = c[i] - 1;
      if (s[i] == 'W') l[i + 1] = l[i] + 1;
      if (s[i] == 'S') l[i + 1] = l[i] - 1;
      if (s[i] == 'W' || s[i] == 'S') c[i + 1] = c[i];
      if (s[i] == 'D' || s[i] == 'A') l[i + 1] = l[i];
      minx = min(minx, l[i + 1]);
      miny = min(miny, c[i + 1]);
      maxx = max(maxx, l[i + 1]);
      maxy = max(maxy, c[i + 1]);
    }
    long long ans = (maxx - minx + 1) * (maxy - miny + 1);
    if (check(l, s.size()) && maxx - minx + 1 > 2)
      ans = min(ans, (maxx - minx) * (maxy - miny + 1));
    if (check(c, s.size()) && maxy - miny + 1 > 2)
      ans = min(ans, (maxx - minx + 1) * (maxy - miny));
    cout << ans << "\n";
  }
  return 0;
}
