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 long long MXN = 2e5 + 10;
long long q, n, now, ans;
string s, dir = "WASD";
vector<long long> dr = {-1, 0, 1, 0}, dc = {0, -1, 0, 1};
long long rmin[MXN], rmax[MXN], cmin[MXN], cmax[MXN], rpos[MXN], cpos[MXN],
    idx[MXN];
void solve() {
  cin >> s;
  n = s.size(), ans = 1e17;
  cpos[0] = rpos[0] = cpos[n + 1] = rpos[n + 1] = 0;
  cmin[0] = rmin[0] = cmin[n + 1] = rmin[n + 1] = 0;
  cmax[0] = rmax[0] = cmax[n + 1] = rmax[n + 1] = 0;
  for (int i = 0; i < n; i++) {
    now = idx[s[i]];
    rpos[i + 1] = rpos[i] + dr[now], cpos[i + 1] = cpos[i] + dc[now];
    rmin[i + 1] = min(rmin[i], rpos[i + 1]),
             rmax[i + 1] = max(rmax[i], rpos[i + 1]);
    cmin[i + 1] = min(cmin[i], cpos[i + 1]),
             cmax[i + 1] = max(cmax[i], cpos[i + 1]);
  }
  for (int i = n; ~i; i--) {
    for (int j = 0; j < 4; j++) {
      long long nowr = rpos[i] + dr[j], nowc = cpos[i] + dc[j];
      long long minr = rmin[i], maxr = rmax[i], minc = cmin[i], maxc = cmax[i];
      minr = min(minr, nowr), maxr = max(maxr, nowr);
      minc = min(minc, nowc), maxc = max(maxc, nowc);
      long long outr = nowr - rpos[i + 1], outc = nowc - cpos[i + 1];
      minr = min(minr, rmin[i + 1] + outr),
      maxr = max(maxr, rmax[i + 1] + outr);
      minc = min(minc, cmin[i + 1] + outc),
      maxc = max(maxc, cmax[i + 1] + outc);
      ans = min(ans, (maxr - minr + 1) * (maxc - minc + 1));
    }
    if (i > 0) {
      long long nw = idx[s[i - 1]] ^ 2;
      rpos[i] = rpos[i + 1] + dr[nw], cpos[i] = cpos[i + 1] + dc[nw];
      rmin[i] = min(rmin[i + 1], rpos[i]), rmax[i] = max(rmax[i + 1], rpos[i]);
      cmin[i] = min(cmin[i + 1], cpos[i]), cmax[i] = max(cmax[i + 1], cpos[i]);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < 4; i++) idx[dir[i]] = i;
  cin >> q;
  while (q--) solve();
  return 0;
}
