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 maxn = 2e6 + 50;
char s[maxn];
struct date {
  long long x, y;
} a[maxn];
int main() {
  long long t;
  scanf("%I64d", &t);
  while (t--) {
    scanf("%s", s);
    long long max = 0, mix = 0, may = 0, miy = 0;
    long long cax1 = 0, cix1 = 0, cay1 = 0, ciy1 = 0;
    long long len = strlen(s);
    long long x = 0, y = 0;
    for (long long i = 0; i < len; i++) {
      if (s[i] == 'W') {
        y++;
      } else if (s[i] == 'S') {
        y--;
      } else if (s[i] == 'D') {
        x++;
      } else if (s[i] == 'A') {
        x--;
      }
      a[i].x = x;
      a[i].y = y;
      if (x > max) {
        max = x;
        cax1 = i;
      }
      if (x < mix) {
        mix = x;
        cix1 = i;
      }
      if (y > may) {
        may = y;
        cay1 = i;
      }
      if (y < miy) {
        miy = y;
        ciy1 = i;
      }
    }
    if (len == 1) {
      printf("2\n");
      continue;
    }
    long long ans = (max - mix + 1) * (may - miy + 1);
    long long flag = 1;
    for (long long i = cix1 + 1; i < len; i++) {
      if (a[i].x == max) {
        flag = 0;
        break;
      }
    }
    if (flag && max - mix > 1) {
      ans = min(ans, (max - mix) * (may - miy + 1));
    }
    flag = 1;
    for (long long i = cax1; i < len; i++) {
      if (a[i].x == mix) {
        flag = 0;
        break;
      }
    }
    if (flag && max - mix > 1) {
      ans = min(ans, (max - mix) * (may - miy + 1));
    }
    flag = 1;
    for (long long i = ciy1 + 1; i < len; i++) {
      if (a[i].y == may) {
        flag = 0;
        break;
      }
    }
    if (flag && may - miy > 1) {
      ans = min(ans, (max - mix + 1) * (may - miy));
    }
    flag = 1;
    for (long long i = cay1 + 1; i < len; i++) {
      if (a[i].y == miy) {
        flag = 0;
        break;
      }
    }
    if (flag && may - miy > 1) {
      ans = min(ans, (max - mix + 1) * (may - miy));
    }
    printf("%I64d\n", ans);
  }
}
