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 INF = 1e9 + 7;
const int mod = 1e9 + 7;
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
void max_self(int &a, int b) { a = max(a, b); }
void min_self(int &a, int b) { a = min(a, b); }
void min_self(long long int &a, long long int b) { a = min(a, b); }
int calc(vector<int> &a) {
  int lastMax = -1, firstMin, pt = *max_element((a).begin(), (a).end()),
      pt2 = *min_element((a).begin(), (a).end());
  int n = a.size();
  for (int i = 0; i < n; i++)
    if (a[i] == pt) lastMax = i;
  for (int i = n - 1; i >= 0; i--)
    if (a[i] == pt2) firstMin = i;
  if (firstMin > lastMax) return max(2, pt - pt2);
  return pt - pt2 + 1;
}
int flip(vector<int> &a) {
  int ans = calc(a);
  for (int i = 0; i < a.size(); i++) a[i] = -a[i];
  min_self(ans, calc(a));
  return ans;
}
int main() {
  int tc = nxt();
  while (tc--) {
    string s;
    cin >> s;
    int n = s.length();
    vector<int> a, b;
    a.push_back(0);
    b.push_back(0);
    for (int i = 0; i < n; i++) {
      if (s[i] == 'A') a.push_back(-1);
      if (s[i] == 'D') a.push_back(1);
      if (s[i] == 'S') b.push_back(-1);
      if (s[i] == 'W') b.push_back(1);
    }
    for (int i = 1; i < a.size(); i++) a[i] += a[i - 1];
    for (int i = 1; i < b.size(); i++) b[i] += b[i - 1];
    int oh = a.size() ? *max_element((a).begin(), (a).end()) -
                            *min_element((a).begin(), (a).end()) + 1
                      : 1;
    int ov = b.size() ? *max_element((b).begin(), (b).end()) -
                            *min_element((b).begin(), (b).end()) + 1
                      : 1;
    int h = a.size() ? flip(a) : 1;
    int v = b.size() ? flip(b) : 1;
    printf("%lld\n", min(1LL * oh * v, 1LL * h * ov));
  }
  return 0;
}
