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;
template <class T>
inline bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
inline bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
const int nax = 2e5 + 10;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
int xp[nax], yp[nax], xin[nax], xax[nax], yin[nax], yax[nax];
void solve() {
  string S;
  cin >> S;
  int N = (int)S.length();
  int x = 0, y = 0;
  for (int i = 0; i < N; ++i) {
    if (S[i] == 'A') --x;
    if (S[i] == 'D') ++x;
    if (S[i] == 'W') ++y;
    if (S[i] == 'S') --y;
    xp[i + 1] = x;
    yp[i + 1] = x;
    xin[i + 1] = min(xin[i], x);
    xax[i + 1] = max(xax[i], x);
    yin[i + 1] = min(yin[i], y);
    yax[i + 1] = max(yax[i], y);
  }
  long long area = (xax[N] - xin[N] + 1) * 1LL * (yax[N] - yin[N] + 1);
  int max_x = x, min_x = x, max_y = y, min_y = y;
  for (int i = N - 1; i >= 0; --i) {
    if (S[i] == 'W') --y;
    if (S[i] == 'A') ++x;
    if (S[i] == 'S') ++y;
    if (S[i] == 'D') --x;
    uax(max_x, x);
    uax(max_y, y);
    uin(min_x, x);
    uin(min_y, y);
    for (int j = 0; j < 4; ++j) {
      long long xx =
          max(xax[i] + dx[j], max_x) - min(xin[i] + dx[j], min_x) + 1LL;
      long long yy =
          max(yax[i] + dy[j], max_y) - min(yin[i] + dy[j], min_y) + 1LL;
      uin(area, xx * yy);
    }
  }
  cout << area << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) solve();
}
