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;
namespace ___ {
int _ = 0;
void testcase() { cout << "Case " << (++_) << ": "; }
}  // namespace ___
template <class T>
inline bool scan(T& ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' and (c < '0' or c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' and c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <class T>
inline void out(T x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
using namespace ___;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos(-1.0);
const double eps = 1e-7;
const int mod = 998244353;
const int maxn = 3e6 + 50;
const int maxm = 7e6 + 10;
const int N = 2e5 + 5;
char a[maxn];
int get(char ch) {
  if (ch == 'W') return 0;
  if (ch == 'S') return 1;
  if (ch == 'A') return 2;
  return 3;
}
int dd[] = {1, -1};
long long get(long long le, long long ri) { return abs(le - ri) + 1LL; }
int use[maxn], mx[maxn], mi[maxn], cc;
pair<long long, long long> solve() {
  if (cc == 0) return pair<long long, long long>(1, 1);
  if (cc == 1) return pair<long long, long long>(2, 2);
  mx[cc + 1] = mi[cc + 1] = 0;
  for (int i = cc; i; --i) {
    if (use[i] == 1) {
      mx[i] = mx[i + 1] + 1;
      mi[i] = mi[i + 1] + 1;
    } else {
      mx[i] = mx[i + 1] - 1;
      mi[i] = mi[i + 1] - 1;
    }
    mx[i] = max(mx[i], 0);
    mi[i] = min(mi[i], 0);
  }
  long long w = INF, now = 0;
  long long u = 0, v = 0;
  for (int i = 1; i <= cc; ++i) {
    now += use[i];
    u = max(u, now);
    v = min(v, now);
    w = min(w, get(max(u, now + 1 + mx[i + 1]), min(v, now + 1 + mi[i + 1])));
    w = min(w, get(max(u, now - 1 + mx[i + 1]), min(v, now - 1 + mi[i + 1])));
  }
  return pair<long long, long long>(w, get(mx[1], mi[1]));
}
void work() {
  cin >> a + 1;
  int n = strlen(a + 1);
  cc = 0;
  for (int i = 1; i <= n; ++i) {
    int d = get(a[i]);
    if (d < 2) {
      use[++cc] = d ? -1 : 1;
    }
  }
  pair<long long, long long> g = solve();
  cc = 0;
  for (int i = 1; i <= n; ++i) {
    int d = get(a[i]);
    if (d >= 2) use[++cc] = (d == 2) ? 1 : -1;
  }
  pair<long long, long long> q = solve();
  long long ans =
      min(g.first * q.second, min(g.second * q.second, g.second * q.first));
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (; t--;) work();
  return 0;
}
