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 mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const long long MAX = 200010;
const long double EPS = 1e-9;
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t) {
    t--;
    string s;
    cin >> s;
    long long n = s.size();
    vector<vector<long long> > rui(n + 1, vector<long long>(4));
    long long nowx = 0, nowy = 0;
    for (long long i = (0); i < (n); i++) {
      if (s[i] == 'D') {
        nowx++;
      } else if (s[i] == 'S') {
        nowy--;
      } else if (s[i] == 'A') {
        nowx--;
      } else {
        nowy++;
      }
      rui[i + 1][0] = max(rui[i][0], nowx);
      rui[i + 1][1] = min(rui[i][1], nowy);
      rui[i + 1][2] = min(rui[i][2], nowx);
      rui[i + 1][3] = max(rui[i][3], nowy);
    }
    long long X = rui[n][0] - rui[n][2] + 1;
    long long Y = rui[n][3] - rui[n][1] + 1;
    long long ans = X * Y;
    vector<vector<long long> > rm(n + 1, vector<long long>(4));
    rm[n][0] = nowx;
    rm[n][1] = nowy;
    rm[n][2] = nowx;
    rm[n][3] = nowy;
    for (long long i = n - 1; i >= 0; i--) {
      if (s[i] == 'D') {
        nowx--;
      } else if (s[i] == 'S') {
        nowy++;
      } else if (s[i] == 'A') {
        nowx++;
      } else {
        nowy--;
      }
      rm[i][0] = max(rm[i + 1][0], nowx);
      rm[i][1] = min(rm[i + 1][1], nowy);
      rm[i][2] = min(rm[i + 1][2], nowx);
      rm[i][3] = max(rm[i + 1][3], nowy);
    }
    for (long long i = (0); i < (n); i++) {
      if (s[i] == 'D') {
        nowx++;
      } else if (s[i] == 'S') {
        nowy--;
      } else if (s[i] == 'A') {
        nowx--;
      } else {
        nowy++;
      }
      if (rui[n][0] - rui[i + 1][0] > 0) {
        if (rm[i + 1][2] > rui[i + 1][2]) ans = min(ans, X * Y - Y);
      }
      if (rui[n][1] - rui[i + 1][1] < 0) {
        if (rm[i + 1][3] < rui[i + 1][3]) ans = min(ans, X * Y - X);
      }
      if (rui[n][2] - rui[i + 1][2] < 0) {
        if (rm[i + 1][0] < rui[i + 1][0]) ans = min(ans, X * Y - Y);
      }
      if (rui[n][3] - rui[i + 1][3] > 0) {
        if (rm[i + 1][1] > rui[i + 1][1]) ans = min(ans, X * Y - X);
      }
    }
    cout << ans << '\n';
  }
}
