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 N = 2e6 + 7, mod = 1e9 + 7;
string s;
long long licz(string tmp) {
  pair<int, int> cur;
  long long mix = 0;
  long long max = 0;
  long long miy = 0;
  long long may = 0;
  for (int i = 0; i < tmp.size(); i++) {
    if (tmp[i] == 'W') cur.first++;
    if (tmp[i] == 'A') cur.second--;
    if (tmp[i] == 'S') cur.first--;
    if (tmp[i] == 'D') cur.second++;
    if (max < cur.first) max = cur.first;
    if (may < cur.second) may = cur.second;
    if (mix > cur.first) mix = cur.first;
    if (miy > cur.second) miy = cur.second;
  }
  long long re = (max - mix + 1) * (may - miy + 1);
  return re;
}
void solve() {
  cin >> s;
  long long ans = licz(s);
  pair<int, int> poc = {0, 0};
  pair<int, int> max = {0, 0};
  pair<int, int> mix = {0, 0};
  pair<int, int> miy = {0, 0};
  pair<int, int> may = {0, 0};
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == 'W') poc.first++;
    if (s[i] == 'A') poc.second--;
    if (s[i] == 'S') poc.first--;
    if (s[i] == 'D') poc.second++;
    if (max.first < poc.first) max = {poc.first, i};
    if (may.first < poc.second) may = {poc.second, i};
    if (mix.first > poc.first) mix = {poc.first, i};
    if (miy.first > poc.second) miy = {poc.second, i};
  }
  string re = s;
  re.insert(re.begin() + mix.second, 'W');
  ans = min(ans, licz(re));
  re = s;
  re.insert(re.begin() + max.second, 'S');
  ans = min(ans, licz(re));
  re = s;
  re.insert(re.begin() + miy.second, 'D');
  ans = min(ans, licz(re));
  re = s;
  re.insert(re.begin() + may.second, 'A');
  ans = min(ans, licz(re));
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  int qwe;
  cin >> qwe;
  while (qwe--) {
    solve();
  }
}
