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 ll = long long;
using namespace std;
long long f(vector<long long> arr, long long depth = 1) {
  vector<long long> pref = arr;
  for (long long i = 1; i < arr.size(); i++) {
    pref[i] += pref[i - 1];
  }
  long long mx = 0, mn = 0;
  for (auto x : pref) {
    mx = max(mx, x);
    mn = min(mn, x);
  }
  long long ind = 0;
  for (long long i = 0; i < arr.size(); i++) {
    if (pref[i] == mx) {
      ind = i;
      break;
    }
  }
  if (depth == 0) {
    return mx + abs(mn) + 1;
  }
  vector<long long> na;
  for (long long i = 0; i < ind; i++) {
    na.push_back(arr[i]);
  }
  na.push_back(-1);
  for (long long i = ind; i < arr.size(); i++) na.push_back(arr[i]);
  return f(na, 0);
}
void sol() {
  string s;
  cin >> s;
  vector<long long> x, y;
  for (auto el : s) {
    if (el == 'W') {
      y.push_back(1);
    } else if (el == 'S') {
      y.push_back(-1);
    } else if (el == 'A') {
      x.push_back(-1);
    } else {
      x.push_back(1);
    }
  }
  long long n = f(x, 0), m = f(y, 0);
  long long n1 = f(x), m1 = f(y);
  for (auto &el : x) {
    el *= -1;
  }
  for (auto &el : y) {
    el *= -1;
  }
  n1 = min(n1, f(x));
  m1 = min(m1, f(y));
  cout << min(n1 * m, n * m1) << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    sol();
  }
}
