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 double PI = 3.14159265358979323846;
const double eps = (1e-4);
int dcmp(long double x, long double y) {
  if (abs(x - y) < eps) return 0;
  if (x > y) return 1;
  return -1;
}
string s;
enum { H, V };
map<int, vector<int> > m[2];
void pre() {
  int v = 0, h = 0;
  m[V][v].push_back(-1);
  m[H][h].push_back(-1);
  for (int i = 0; i < (int)(s).size(); ++i) {
    if (s[i] == 'W') {
      v++;
      m[V][v].push_back(i);
    } else if (s[i] == 'S') {
      v--;
      m[V][v].push_back(i);
    } else if (s[i] == 'D') {
      h++;
      m[H][h].push_back(i);
    } else {
      h--;
      m[H][h].push_back(i);
    }
  }
}
long long calc(int id) {
  return 1LL * (m[id].rbegin()->first - m[id].begin()->first + 1);
}
long long f(const pair<int, vector<int> > &p1,
            const pair<int, vector<int> > &p2) {
  if ((p1.first == 0 && p2.first == 0)) return 1;
  if (abs(p1.first - p2.first) == 1) return 2;
  if (p1.second[0] < p2.second.back()) {
    return abs(p1.first - p2.first) + 1;
  }
  return abs(p1.first - p2.first);
}
long long solve(int id) {
  pair<int, vector<int> > p1 = *m[id].begin();
  pair<int, vector<int> > p2 = *m[id].rbegin();
  long long ret = calc(!id) * f(p1, p2);
  ret = min(ret, calc(!id) * f(p2, p1));
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    for (int i = 0; i < 2; ++i) m[i].clear();
    cin >> s;
    pre();
    long long ans = 1LL * calc(H) * calc(V);
    ans = min(ans, solve(H));
    ans = min(ans, solve(V));
    cout << ans << '\n';
  }
  return 0;
}
