Codeforces Round #742 (Div. 2)

A. Domino Disaster

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Alice has a grid with 22 rows and nn columns. She fully covers the grid using nn dominoes of size 1×21×2 — Alice may place them vertically or horizontally, and each cell should be covered by exactly one domino.

Now, she decided to show one row of the grid to Bob. Help Bob and figure out what the other row of the grid looks like!

Input

The input consists of multiple test cases. The first line contains an integer tt (1≤t≤50001≤t≤5000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer nn (1≤n≤1001≤n≤100) — the width of the grid.

The second line of each test case contains a string ss consisting of nn characters, each of which is either L, R, U, or D, representing the left, right, top, or bottom half of a domino, respectively (see notes for better understanding). This string represents one of the rows of the grid.

Additional constraint on the input: each input corresponds to at least one valid tiling.

Output

For each test case, output one string — the other row of the grid, using the same format as the input string. If there are multiple answers, print any.

Example

input

Copy

4

1

U

2

LR

5

LRDLR

6

UUUUUU

output

Copy

D

LR

LRULR

DDDDDD

Note

In the first test case, Alice shows Bob the top row, the whole grid may look like

In the second test case, Alice shows Bob the bottom row, the whole grid may look like:

In the third test case, Alice shows Bob the bottom row, the whole grid may look like:

In the fourth test case, Alice shows Bob the top row, the whole grid may look like:

题目挺简单的，直接上代码。

#include <iostream>
using namespace std;
#define ll long long
char s[333];
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    scanf("%s", s + 1);
    for (int i = 1; i <= n; i++) {
      if (s[i] == 'U') {
        printf("D");
        continue;
      }
      if (s[i] == 'D') {
        printf("U");
        continue;
      }
      printf("%c", s[i]);
    }
    printf("\n");
  }
}