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 const maxn = 5e5 + 5;
int const td[4] = {-1, 0, 1, 0};
int const tc[4] = {0, -1, 0, 1};
struct grid {
  int xM, yM, xm, ym;
};
int test, n;
pair<int, int> p[maxn];
grid a[maxn], b[maxn];
pair<int, int> go(pair<int, int> pos, char com) {
  int k;
  switch (com) {
    case 'W': {
      k = 0;
      break;
    }
    case 'A': {
      k = 1;
      break;
    }
    case 'S': {
      k = 2;
      break;
    }
    case 'D': {
      k = 3;
      break;
    }
  }
  return pair<int, int>(pos.first + td[k], pos.second + tc[k]);
}
grid update(grid a, pair<int, int> pos) {
  a.xM = max(a.xM, pos.first);
  a.xm = min(a.xm, pos.first);
  a.yM = max(a.yM, pos.second);
  a.ym = min(a.ym, pos.second);
  return a;
}
grid mov(grid d, pair<int, int> pos, int k) {
  d.xM += td[k];
  d.xm += td[k];
  d.yM += tc[k];
  d.ym += tc[k];
  pos.first += td[k];
  pos.second += tc[k];
  d.xM = max(d.xM, pos.first);
  d.xm = min(d.xm, pos.first);
  d.yM = max(d.yM, pos.second);
  d.ym = min(d.ym, pos.second);
  return d;
}
long long S(grid a, grid b) {
  a.xM = max(a.xM, b.xM);
  a.xm = min(a.xm, b.xm);
  a.yM = max(a.yM, b.yM);
  a.ym = min(a.ym, b.ym);
  return 1LL * (a.xM - a.xm + 1) * (a.yM - a.ym + 1);
}
void sol() {
  string s;
  cin >> s;
  n = s.size();
  p[0] = pair<int, int>(0, 0);
  for (int i = 0; i < n; i++) p[i + 1] = go(p[i], s[i]);
  a[0] = {0, 0, 0, 0};
  for (int i = 1; i <= n; i++) a[i] = update(a[i - 1], p[i]);
  b[n] = {p[n].first, p[n].second, p[n].first, p[n].second};
  for (int i = n - 1; i >= 1; i--) b[i] = update(b[i + 1], p[i]);
  long long ans = S(a[0], b[1]);
  for (int i = 2; i <= s.size(); i++)
    for (int j = 0; j <= 3; j++)
      ans = min(ans, S(a[i - 1], mov(b[i], p[i - 1], j)));
  cout << ans << endl;
}
int main() {
  scanf("%d", &test);
  while (test--) sol();
  return 0;
}
