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, n;
string s;
int p1[200005], p2[200005], c1, c2, u, d, l, r;
void init() {
  c1 = 0, c2 = 0;
  n = s.length();
  s = "*" + s;
  for (int i = 1; i <= n; i++) {
    if (s[i] == 'W') {
      p1[c1 + 1] = p1[c1] + 1;
      c1++;
    } else if (s[i] == 'S') {
      p1[c1 + 1] = p1[c1] - 1;
      c1++;
    } else if (s[i] == 'A') {
      p2[c2 + 1] = p2[c2] + 1;
      c2++;
    } else {
      p2[c2 + 1] = p2[c2] - 1;
      c2++;
    }
  }
  u = *max_element(p1, p1 + c1 + 1);
  d = *min_element(p1, p1 + c1 + 1);
  l = *max_element(p2, p2 + c2 + 1);
  r = *min_element(p2, p2 + c2 + 1);
}
long long ans;
void slove() {
  ans = 1ll * (u - d + 1) * (l - r + 1);
  int a, b, x, y;
  for (int i = 0; i <= c1; i++) {
    if (p1[i] == u) a = i;
    if (p1[c1 - i] == d) b = c1 - i;
    if (p1[i] == d) x = i;
    if (p1[c1 - i] == u) y = c1 - i;
  }
  if (a + 1 < b || x + 1 < y) ans = min(ans, 1ll * (l - r + 1) * (u - d));
  for (int i = 0; i <= c2; i++) {
    if (p2[i] == l) a = i;
    if (p2[c2 - i] == r) b = c2 - i;
    if (p2[i] == r) x = i;
    if (p2[c2 - i] == l) y = c2 - i;
  }
  if (a + 1 < b || x + 1 < y) ans = min(ans, 1ll * (l - r) * (u - d + 1));
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> s;
    init();
    slove();
    cout << ans << endl;
  }
}
