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 long long mod = 1e9 + 7;
const int maxn = 2e5 + 5;
int n, m, k, T;
struct node {
  long long w, a, s, d;
} pre[maxn], suf[maxn];
char c[maxn];
void solve() {
  scanf("%s", c + 1);
  n = strlen(c + 1);
  int x = 0, y = 0, w = 0, a = 0, s = 0, d = 0;
  for (int i = 1; i <= n; i++) {
    if (c[i] == 'W') ++y;
    if (c[i] == 'S') --y;
    if (c[i] == 'A') --x;
    if (c[i] == 'D') ++x;
    w = max(w, y);
    a = min(a, x);
    s = min(s, y);
    d = max(d, x);
    pre[i].w = w - y, pre[i].a = a - x, pre[i].s = s - y, pre[i].d = d - x;
  }
  x = 0, y = 0, w = 0, a = 0, s = 0, d = 0;
  for (int i = n; i >= 1; i--) {
    if (c[i] == 'W') --y;
    if (c[i] == 'S') ++y;
    if (c[i] == 'A') ++x;
    if (c[i] == 'D') --x;
    w = max(w, y);
    a = min(a, x);
    s = min(s, y);
    d = max(d, x);
    suf[i].w = w - y, suf[i].a = a - x, suf[i].s = s - y, suf[i].d = d - x;
  }
  long long ans = 1ll * (suf[1].d - suf[1].a + 1) * (suf[1].w - suf[1].s + 1);
  for (int i = 1; i <= n; i++) {
    ans = min(ans, 1ll *
                       (max(pre[i].w, suf[i + 1].w + 1) -
                        min(pre[i].s, suf[i + 1].s + 1) + 1) *
                       1ll *
                       (max(pre[i].d, suf[i + 1].d) -
                        min(pre[i].a, suf[i + 1].a) + 1));
    ans = min(ans, 1ll *
                       (max(pre[i].w, suf[i + 1].w - 1) -
                        min(pre[i].s, suf[i + 1].s - 1) + 1) *
                       1ll *
                       (max(pre[i].d, suf[i + 1].d) -
                        min(pre[i].a, suf[i + 1].a) + 1));
    ans = min(
        ans,
        1ll * (max(pre[i].w, suf[i + 1].w) - min(pre[i].s, suf[i + 1].s) + 1) *
            1ll *
            (max(pre[i].d, suf[i + 1].d + 1) - min(pre[i].a, suf[i + 1].a + 1) +
             1));
    ans = min(
        ans,
        1ll * (max(pre[i].w, suf[i + 1].w) - min(pre[i].s, suf[i + 1].s) + 1) *
            1ll *
            (max(pre[i].d, suf[i + 1].d - 1) - min(pre[i].a, suf[i + 1].a - 1) +
             1));
  }
  printf("%lld\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}
