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 N = 2e5 + 15;
char s[N];
int maxi[N], mini[N];
pair<int, int> getLen(vector<int> &in) {
  int ret = 1e9;
  int n = in.size();
  maxi[n] = mini[n] = 0;
  int tmpMaxi = 0, tmpMini = 0;
  int cur = 0;
  for (int i = n - 1; i >= 0; --i) {
    int tmp = in[i];
    maxi[i] = max(tmp, tmp + maxi[i + 1]);
    mini[i] = min(tmp, tmp + mini[i + 1]);
    maxi[i] = max(maxi[i], 0);
    mini[i] = min(mini[i], 0);
  }
  ret = maxi[0] - mini[0] + 1;
  cur = 0;
  for (int i = 0; i <= in.size(); ++i) {
    for (int j = -1; j <= 1; j += 2) {
      int tmpCur = cur + j;
      int cmaxi = max(tmpCur + maxi[i], tmpMaxi);
      int cmini = min(tmpCur + mini[i], tmpMini);
      ret = min(ret, cmaxi - cmini + 1);
    }
    if (i == in.size()) {
      break;
    } else {
      cur += in[i];
      tmpMaxi = max(tmpMaxi, cur);
      tmpMini = min(tmpMini, cur);
    }
  }
  return make_pair(maxi[0] - mini[0] + 1, ret);
}
void solve() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%s", s);
    int n = strlen(s);
    vector<int> X, Y;
    for (int i = 0; i < n; ++i) {
      if (s[i] == 'W') {
        Y.push_back(1);
      } else if (s[i] == 'S') {
        Y.push_back(-1);
      } else if (s[i] == 'A') {
        X.push_back(1);
      } else if (s[i] == 'D') {
        X.push_back(-1);
      } else {
        assert(0);
      }
    }
    pair<int, int> x = getLen(X);
    pair<int, int> y = getLen(Y);
    long long ret = (long long)x.first * y.second;
    ret = min(ret, (long long)x.second * y.first);
    printf("%lld\n", ret);
  }
}
int main() { solve(); }
