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 = 200005;
const string lets = "AWDS";
const int Maxd = 4;
const int dy[Maxd] = {0, -1, 0, 1};
const int dx[Maxd] = {-1, 0, 1, 0};
int t;
string s;
int mnx[Maxn], mny[Maxn], mxx[Maxn], mxy[Maxn];
long long Area(long long mnx, long long mny, long long mxx, long long mxy) {
  return (mxx - mnx + 1ll) * (mxy - mny + 1ll);
}
int main() {
  scanf("%d", &t);
  while (t--) {
    cin >> s;
    mnx[s.length()] = mny[s.length()] = mxx[s.length()] = mxy[s.length()] = 0;
    for (int i = int(s.length()) - 1; i >= 0; i--) {
      int ind = lets.find(s[i]);
      mnx[i] = min(0, mnx[i + 1] + dx[ind]);
      mxx[i] = max(0, mxx[i + 1] + dx[ind]);
      mny[i] = min(0, mny[i + 1] + dy[ind]);
      mxy[i] = max(0, mxy[i + 1] + dy[ind]);
    }
    long long res = Area(mnx[0], mny[0], mxx[0], mxy[0]);
    long long curx = 0, cury = 0;
    long long X1, X2;
    X1 = X2 = curx;
    long long Y1, Y2;
    Y1 = Y2 = cury;
    for (int i = 0; i < s.length(); i++) {
      int ind = lets.find(s[i]);
      curx += dx[ind], cury += dy[ind];
      X1 = min(X1, curx);
      X2 = max(X2, curx);
      Y1 = min(Y1, cury);
      Y2 = max(Y2, cury);
      for (int j = 0; j < Maxd; j++) {
        long long gX1 = min(X1, curx + dx[j] + mnx[i + 1]);
        long long gX2 = max(X2, curx + dx[j] + mxx[i + 1]);
        long long gY1 = min(Y1, cury + dy[j] + mny[i + 1]);
        long long gY2 = max(Y2, cury + dy[j] + mxy[i + 1]);
        res = min(res, Area(gX1, gY1, gX2, gY2));
      }
    }
    printf("%I64d\n", res);
  }
  return 0;
}
