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;
bool ok(vector<long long> v) {
  int n = v.size();
  for (int i = 0; i < n; ++i) v[i] = 2 * v[i] - 1;
  int x = 0;
  int mi = 0;
  int ma = 0;
  for (int i = 0; i < n; ++i) {
    x += v[i];
    if (mi > x) mi = x;
    if (ma < x) ma = x;
  }
  if (ma - mi < 2) return false;
  vector<long long> u(2, 0);
  x = 0;
  int i = 0;
  while (x != ma && x != mi) {
    x += v[i];
    ++i;
  }
  int fi = x;
  int se = mi + ma - fi;
  while (x != se) {
    x += v[i];
    ++i;
  }
  while (i < n) {
    x += v[i];
    ++i;
    if (x == fi) return false;
  }
  return true;
}
long long f(vector<long long> v) {
  int n = v.size();
  for (int i = 0; i < n; ++i) v[i] = 2 * v[i] - 1;
  int x = 0;
  int mi = 0;
  int ma = 0;
  for (int i = 0; i < n; ++i) {
    x += v[i];
    if (mi > x) mi = x;
    if (ma < x) ma = x;
  }
  return ma - mi + 1;
}
long long solve() {
  string s;
  cin >> s;
  int n = s.length();
  vector<long long> v(n);
  for (int i = 0; i < n; ++i) {
    if (s[i] == 'W') v[i] = 0;
    if (s[i] == 'S') v[i] = 1;
    if (s[i] == 'D') v[i] = 2;
    if (s[i] == 'A') v[i] = 3;
  }
  vector<long long> a, b;
  for (int i = 0; i < n; ++i) {
    if (v[i] < 2)
      a.push_back(v[i]);
    else
      b.push_back(v[i] - 2);
  }
  long long lenx = f(b);
  long long leny = f(a);
  long long res = lenx * leny;
  if (ok(b)) res = min(res, (lenx - 1) * leny);
  if (ok(a)) res = min(res, lenx * (leny - 1));
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) cout << solve() << "\n";
  return 0;
}
