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 int MOD = 1e9 + 7;
const int INF = 2e9 + 5;
const int MAXN = 200010;
int T, n;
string s;
int width[MAXN], height[MAXN];
int maxwidth[MAXN], minwidth[MAXN];
int maxheight[MAXN], minheight[MAXN];
int best(int a[], int mx[], int mn[], char le, char ri) {
  for (int i = 1; i <= n; i++) {
    a[i] = a[i - 1];
    if (s[i - 1] == ri) ++a[i];
    if (s[i - 1] == le) --a[i];
  }
  for (int i = n; i >= 0; i--) {
    mx[i] = max(mx[i + 1], a[i]);
    mn[i] = min(mn[i + 1], a[i]);
  }
  int maxlen = 0, minlen = 0;
  int bestlen = mx[0] - mn[0];
  for (int i = 1; i < n; i++) {
    maxlen = max(maxlen, a[i]);
    minlen = min(minlen, a[i]);
    int newmin, newmax;
    newmax = max(mx[i + 1] - 1, maxlen);
    newmin = min(mn[i + 1] - 1, min(a[i] - 1, minlen));
    bestlen = min(bestlen, newmax - newmin);
    newmax = max(mx[i + 1] + 1, max(a[i] + 1, maxlen));
    newmin = min(mn[i + 1] + 1, minlen);
    bestlen = min(bestlen, newmax - newmin);
  }
  return bestlen;
}
void solve() {
  cin >> s;
  n = s.length();
  for (int i = 0; i <= n + 2; i++) {
    height[i] = width[i] = 0;
    maxwidth[i] = maxheight[i] = -INF;
    minwidth[i] = minheight[i] = INF;
  }
  long long besth = best(height, maxheight, minheight, 'S', 'W');
  long long bestw = best(width, maxwidth, minwidth, 'A', 'D');
  long long oldw = maxwidth[0] - minwidth[0];
  long long oldh = maxheight[0] - minheight[0];
  long long ans = (oldw + 1) * (oldh + 1);
  ans = min(ans, (oldw + 1) * (besth + 1));
  ans = min(ans, (oldh + 1) * (bestw + 1));
  cout << ans << "\n";
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) {
    solve();
  }
}
