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 INF = (int)1e9;
const int MXN = (int)1e6 + 7;
const int mod = (int)1e9 + 7;
char s[MXN];
int xx[] = {0, 1, 0, -1};
int yy[] = {1, 0, -1, 0};
int dir(char ch) {
  if (ch == 'W')
    return 0;
  else if (ch == 'D')
    return 1;
  else if (ch == 'S')
    return 2;
  else
    return 3;
}
pair<int, int> pos[MXN];
int lmxn[MXN][4], rmxn[MXN][4];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    for (int i = 0; i < 4; i++) rmxn[l + 1][i] = 0;
    int x = 0, y = 0;
    for (int i = 1; i <= l; i++) {
      int d = dir(s[i]);
      x += xx[d];
      y += yy[d];
      pos[i] = {x, y};
      for (int j = 0; j < 4; j++) lmxn[i][j] = lmxn[i - 1][j];
      lmxn[i][0] = max(lmxn[i][0], y);
      lmxn[i][1] = max(lmxn[i][1], x);
      lmxn[i][2] = min(lmxn[i][2], y);
      lmxn[i][3] = min(lmxn[i][3], x);
    }
    long long ans =
        (lmxn[l][0] - lmxn[l][2] + 1) * 1ll * (lmxn[l][1] - lmxn[l][3] + 1);
    x = 0, y = 0;
    for (int i = l; i >= 1; i--) {
      int d = dir(s[i]);
      d = (d + 2) % 4;
      x += xx[d];
      y += yy[d];
      for (int j = 0; j < 4; j++) rmxn[i][j] = rmxn[i + 1][j];
      rmxn[i][0] = max(rmxn[i][0], y);
      rmxn[i][1] = max(rmxn[i][1], x);
      rmxn[i][2] = min(rmxn[i][2], y);
      rmxn[i][3] = min(rmxn[i][3], x);
      int l, r, b, t;
      t = max(rmxn[i][0] - y + 1, lmxn[i - 1][0] - pos[i - 1].second);
      r = max(rmxn[i][1] - x, lmxn[i - 1][1] - pos[i - 1].first);
      d = min(rmxn[i][2] - y + 1, lmxn[i - 1][2] - pos[i - 1].second);
      l = min(rmxn[i][3] - x, lmxn[i - 1][3] - pos[i - 1].first);
      ans = min(ans, (t - d + 1) * 1ll * (r - l + 1));
      t = max(rmxn[i][0] - y, lmxn[i - 1][0] - pos[i - 1].second);
      r = max(rmxn[i][1] - x + 1, lmxn[i - 1][1] - pos[i - 1].first);
      d = min(rmxn[i][2] - y, lmxn[i - 1][2] - pos[i - 1].second);
      l = min(rmxn[i][3] - x + 1, lmxn[i - 1][3] - pos[i - 1].first);
      ans = min(ans, (t - d + 1) * 1ll * (r - l + 1));
      t = max(rmxn[i][0] - y - 1, lmxn[i - 1][0] - pos[i - 1].second);
      r = max(rmxn[i][1] - x, lmxn[i - 1][1] - pos[i - 1].first);
      d = min(rmxn[i][2] - y - 1, lmxn[i - 1][2] - pos[i - 1].second);
      l = min(rmxn[i][3] - x, lmxn[i - 1][3] - pos[i - 1].first);
      ans = min(ans, (t - d + 1) * 1ll * (r - l + 1));
      t = max(rmxn[i][0] - y, lmxn[i - 1][0] - pos[i - 1].second);
      r = max(rmxn[i][1] - x - 1, lmxn[i - 1][1] - pos[i - 1].first);
      d = min(rmxn[i][2] - y, lmxn[i - 1][2] - pos[i - 1].second);
      l = min(rmxn[i][3] - x - 1, lmxn[i - 1][3] - pos[i - 1].first);
      ans = min(ans, (t - d + 1) * 1ll * (r - l + 1));
    }
    printf("%lld\n", ans);
  }
  return 0;
}
