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 maxs = 200000;
const char dbuf[] = "DWAS";
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
char s[maxs + 1];
int xv[maxs + 1];
int yv[maxs + 1];
int lprv[maxs + 1];
int bprv[maxs + 1];
int rprv[maxs + 1];
int tprv[maxs + 1];
int lnxt[maxs + 1];
int bnxt[maxs + 1];
int rnxt[maxs + 1];
int tnxt[maxs + 1];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T;
  cin >> T;
  for (int TN = 0; TN < T; TN++) {
    cin >> s;
    int n = strlen(s);
    xv[0] = 0;
    yv[0] = 0;
    for (int i = 0; i < n; i++) {
      int d = find(dbuf, dbuf + 4, s[i]) - dbuf;
      xv[i + 1] = xv[i] + dx[d];
      yv[i + 1] = yv[i] + dy[d];
    }
    lprv[0] = rprv[0] = xv[0];
    bprv[0] = tprv[0] = yv[0];
    for (int i = 1; i <= n; i++) {
      lprv[i] = min(lprv[i - 1], xv[i]);
      bprv[i] = min(bprv[i - 1], yv[i]);
      rprv[i] = max(rprv[i - 1], xv[i]);
      tprv[i] = max(tprv[i - 1], yv[i]);
    }
    lnxt[n] = rnxt[n] = xv[n];
    bnxt[n] = tnxt[n] = yv[n];
    for (int i = n - 1; i >= 0; i--) {
      lnxt[i] = min(lnxt[i + 1], xv[i]);
      bnxt[i] = min(bnxt[i + 1], yv[i]);
      rnxt[i] = max(rnxt[i + 1], xv[i]);
      tnxt[i] = max(tnxt[i + 1], yv[i]);
    }
    long long ans = 0x7f7f7f7f7f7f7f7fll;
    for (int i = 0; i <= n; i++) {
      for (int d = 0; d < 4; d++) {
        int w =
            max(rprv[i], rnxt[i] + dx[d]) - min(lprv[i], lnxt[i] + dx[d]) + 1;
        int h =
            max(tprv[i], tnxt[i] + dy[d]) - min(bprv[i], bnxt[i] + dy[d]) + 1;
        ans = min(ans, (long long)w * h);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
