#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, k, num;
  scanf("%d %d %d", &n, &m, &k);
  string sol = "";
  for (int i = 0; i < n - 1; i++) sol.append("D");
  for (int i = 0; i < m - 1; i++) sol.append("L");
  string up = "";
  for (int i = 0; i < n - 1; i++) up.append("U");
  string down = "";
  for (int i = 0; i < n - 1; i++) down.append("D");
  for (int i = 0; i < m; i++) {
    if (i % 2)
      sol.append(down);
    else
      sol.append(up);
    sol.append("R");
  }
  num = n - 1 + m - 1 + n * m;
  printf("%d\n", num);
  cout << sol;
}

### Prompt

Create a solution in CPP for the following problem:

Petya has a rectangular Board of size n × m. Initially, k chips are placed on the board, i-th chip is located in the cell at the intersection of sx_i-th row and sy_i-th column.

In one action, Petya can move all the chips to the left, right, down or up by 1 cell.

If the chip was in the (x, y) cell, then after the operation: 

  * left, its coordinates will be (x, y - 1); 
  * right, its coordinates will be (x, y + 1); 
  * down, its coordinates will be (x + 1, y); 
  * up, its coordinates will be (x - 1, y). 



If the chip is located by the wall of the board, and the action chosen by Petya moves it towards the wall, then the chip remains in its current position.

Note that several chips can be located in the same cell.

For each chip, Petya chose the position which it should visit. Note that it's not necessary for a chip to end up in this position.

Since Petya does not have a lot of free time, he is ready to do no more than 2nm actions.

You have to find out what actions Petya should do so that each chip visits the position that Petya selected for it at least once. Or determine that it is not possible to do this in 2nm actions.

Input

The first line contains three integers n, m, k (1 ≤ n, m, k ≤ 200) — the number of rows and columns of the board and the number of chips, respectively.

The next k lines contains two integers each sx_i, sy_i ( 1 ≤ sx_i ≤ n, 1 ≤ sy_i ≤ m) — the starting position of the i-th chip.

The next k lines contains two integers each fx_i, fy_i ( 1 ≤ fx_i ≤ n, 1 ≤ fy_i ≤ m) — the position that the i-chip should visit at least once.

Output

In the first line print the number of operations so that each chip visits the position that Petya selected for it at least once.

In the second line output the sequence of operations. To indicate operations left, right, down, and up, use the characters L, R, D, U respectively.

If the required sequence does not exist, print -1 in the single line.

Examples

Input


3 3 2
1 2
2 1
3 3
3 2


Output


3
DRD

Input


5 4 3
3 4
3 1
3 3
5 3
1 3
1 4


Output


9
DDLUUUURR

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m, k, num;
  scanf("%d %d %d", &n, &m, &k);
  string sol = "";
  for (int i = 0; i < n - 1; i++) sol.append("D");
  for (int i = 0; i < m - 1; i++) sol.append("L");
  string up = "";
  for (int i = 0; i < n - 1; i++) up.append("U");
  string down = "";
  for (int i = 0; i < n - 1; i++) down.append("D");
  for (int i = 0; i < m; i++) {
    if (i % 2)
      sol.append(down);
    else
      sol.append(up);
    sol.append("R");
  }
  num = n - 1 + m - 1 + n * m;
  printf("%d\n", num);
  cout << sol;
}

```