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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    string S;
    cin >> S;
    S = "*" + S;
    int N = (int)S.size();
    vector<int> h(N + 1), v(N + 1);
    for (int i = 1; i <= N; i++) {
      h[i] = h[i - 1];
      v[i] = v[i - 1];
      if (S[i] == 'D') {
        h[i]++;
      }
      if (S[i] == 'A') {
        h[i]--;
      }
      if (S[i] == 'W') {
        v[i]++;
      }
      if (S[i] == 'S') {
        v[i]--;
      }
    }
    vector<int> ml(N + 1), mr(N + 1), mu(N + 1), md(N + 1);
    mr[N] = h[N];
    ml[N] = h[N];
    mu[N] = v[N];
    md[N] = v[N];
    for (int i = N - 1; i >= 0; i--) {
      mr[i] = max(h[i], mr[i + 1]);
      ml[i] = min(h[i], ml[i + 1]);
      mu[i] = max(v[i], mu[i + 1]);
      md[i] = min(v[i], md[i + 1]);
    }
    long long ans = 1LL * (abs(md[0]) + mu[0] + 1) * (abs(ml[0]) + mr[0] + 1);
    int ch = 0, cv = 0;
    int cl = 0, cr = 0, cu = 0, cd = 0;
    for (int i = 0; i < N; i++) {
      long long R = 1LL *
                    (1 + max({cr, ch + 1, mr[i] + 1}) +
                     abs(min({cl, ch + 1, ml[i] + 1}))) *
                    (1 + mu[0] + abs(md[0]));
      long long L = 1LL *
                    (1 + max({cr, ch - 1, mr[i] - 1}) +
                     abs(min({cl, ch - 1, ml[i] - 1}))) *
                    (1 + mu[0] + abs(md[0]));
      long long U = 1LL *
                    (1 + max({cu, cv + 1, mu[i] + 1}) +
                     abs(min({cd, cv + 1, md[i] + 1}))) *
                    (1 + mr[0] + abs(ml[0]));
      long long D = 1LL *
                    (1 + max({cu, cv - 1, mu[i] - 1}) +
                     abs(min({cd, cv - 1, md[i] - 1}))) *
                    (1 + mr[0] + abs(ml[0]));
      ans = min({ans, L, R, U, D});
      if (S[i] == 'D') {
        ch++;
        cr = max(ch, cr);
      }
      if (S[i] == 'A') {
        ch--;
        cl = min(ch, cl);
      }
      if (S[i] == 'W') {
        cv++;
        cu = max(cv, cu);
      }
      if (S[i] == 'S') {
        cv--;
        cd = min(cv, cd);
      }
    }
    cout << ans << '\n';
  }
}
