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 maxn = 2e5 + 10;
int pL[maxn], sL[maxn], pR[maxn], sR[maxn];
int pU[maxn], sU[maxn], pD[maxn], sD[maxn];
char s[maxn];
int main() {
  int t, n;
  cin >> t;
  while (t--) {
    scanf("%s", s);
    n = strlen(s);
    int x = 0, y = 0;
    for (int i = 0; i < n; i++) {
      pL[i + 1] = pL[i], pR[i + 1] = pR[i];
      pU[i + 1] = pU[i], pD[i + 1] = pD[i];
      if (s[i] == 'W' || s[i] == 'S') {
        if (s[i] == 'W')
          y++;
        else
          y--;
        pU[i + 1] = max(pU[i], y);
        pD[i + 1] = min(pD[i], y);
      } else {
        if (s[i] == 'A')
          x--;
        else
          x++;
        pR[i + 1] = max(pR[i], x);
        pL[i + 1] = min(pL[i], x);
      }
    }
    sL[n] = sR[n] = x;
    sU[n] = sD[n] = y;
    for (int i = n - 1; i >= 0; --i) {
      sL[i] = sL[i + 1], sR[i] = sR[i + 1];
      sU[i] = sU[i + 1], sD[i] = sD[i + 1];
      if (s[i] == 'W' || s[i] == 'S') {
        if (s[i] == 'W')
          y--;
        else
          y++;
        sU[i] = max(sU[i + 1], y);
        sD[i] = min(sD[i + 1], y);
      } else {
        if (s[i] == 'A')
          x++;
        else
          x--;
        sR[i] = max(sR[i + 1], x);
        sL[i] = min(sL[i + 1], x);
      }
    }
    x = (pR[n] - pL[n] + 1), y = (pU[n] - pD[n] + 1);
    long long ans = (long long)x * y;
    for (int i = 0; i < n; i++) {
      int L, R, U, D;
      ans = min(ans, (long long)x *
                         ((max(pU[i], sU[i] + 1) - min(pD[i], sD[i] + 1)) + 1));
      ans = min(ans, (long long)x *
                         ((max(pU[i], sU[i] - 1) - min(pD[i], sD[i] - 1)) + 1));
      ans = min(ans, (long long)y *
                         ((max(pR[i], sR[i] + 1) - min(pL[i], sL[i] + 1)) + 1));
      ans = min(ans, (long long)y *
                         ((max(pR[i], sR[i] - 1) - min(pL[i], sL[i] - 1)) + 1));
    }
    cout << ans << endl;
  }
  return 0;
}
