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 + 5;
const int MAXM = 1e4 + 5;
const int MOD = 998244353;
const int INF = 0x3f3f3f3f;
char s[MAXN];
int s1[MAXN], s2[MAXN];
int calcMinLen(int* presum, int n) {
  int ma = -INF, lastmax, mi = INF, firstmin;
  for (int i = 1; i <= n; ++i) {
    if (presum[i] >= ma) {
      ma = presum[i];
      lastmax = i;
    }
    if (presum[i] < mi) {
      mi = presum[i];
      firstmin = i;
    }
  }
  int ret = ma - mi + 1;
  if (lastmax < firstmin) --ret;
  return ret;
}
void init() { s1[1] = s2[1] = 0; }
int main() {
  int _;
  scanf("%d", &_);
  while (_--) {
    init();
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    int n1 = 1, n2 = 1;
    int miw = 0, maw = 0;
    int mih = 0, mah = 0;
    for (int i = 1; i <= n; ++i) {
      if (s[i] == 'A' || s[i] == 'D') {
        ++n1;
        s1[n1] = s1[n1 - 1] + (s[i] == 'A' ? -1 : 1);
        miw = min(miw, s1[n1]);
        maw = max(maw, s1[n1]);
      } else {
        ++n2;
        s2[n2] = s2[n2 - 1] + (s[i] == 'S' ? -1 : 1);
        mih = min(mih, s2[n2]);
        mah = max(mah, s2[n2]);
      }
    }
    long long ans = 1LL * (maw - miw + 1) * (mah - mih + 1);
    if (n1 >= 3) {
      int width = calcMinLen(s1, n1);
      ans = min(ans, 1LL * width * (mah - mih + 1));
      transform(s1 + 1, s1 + n1 + 1, s1 + 1, negate<int>());
      width = calcMinLen(s1, n1);
      ans = min(ans, 1LL * width * (mah - mih + 1));
    }
    if (n2 >= 3) {
      int height = calcMinLen(s2, n2);
      ans = min(ans, 1LL * (maw - miw + 1) * height);
      transform(s2 + 1, s2 + n2 + 1, s2 + 1, negate<int>());
      height = calcMinLen(s2, n2);
      ans = min(ans, 1LL * (maw - miw + 1) * height);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
