Problem:
Score : 600 points

Problem Statement

Snuke is introducing a robot arm with the following properties to his factory:

The robot arm consists of m sections and m+1 joints. The sections are numbered 1, 2, ..., m, and the joints are numbered 0, 1, ..., m. Section i connects Joint i-1 and Joint i. The length of Section i is d_i.

For each section, its mode can be specified individually. There are four modes: L, R, D and U. The mode of a section decides the direction of that section. If we consider the factory as a coordinate plane, the position of Joint i will be determined as follows (we denote its coordinates as (x_i, y_i)):

(x_0, y_0) = (0, 0).

If the mode of Section i is L, (x_{i}, y_{i}) = (x_{i-1} - d_{i}, y_{i-1}).

If the mode of Section i is R, (x_{i}, y_{i}) = (x_{i-1} + d_{i}, y_{i-1}).

If the mode of Section i is D, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} - d_{i}).

If the mode of Section i is U, (x_{i}, y_{i}) = (x_{i-1}, y_{i-1} + d_{i}).

Snuke would like to introduce a robot arm so that the position of Joint m can be matched with all of the N points (X_1, Y_1), (X_2, Y_2), ..., (X_N, Y_N) by properly specifying the modes of the sections.
Is this possible?
If so, find such a robot arm and how to bring Joint m to each point (X_j, Y_j).

Constraints

All values in input are integers.

1 \leq N \leq 1000

-10^9 \leq X_i \leq 10^9

-10^9 \leq Y_i \leq 10^9

Partial Score

In the test cases worth 300 points, -10 \leq X_i \leq 10 and -10 \leq Y_i \leq 10 hold.

Input

Input is given from Standard Input in the following format:

N
X_1 Y_1
X_2 Y_2
:
X_N Y_N

Output

If the condition can be satisfied, follow the following format. If the condition cannot be satisfied, print -1.

m
d_1 d_2 ... d_m
w_1
w_2
:
w_N

m and d_i are the configurations of the robot arm. Refer to the problem statement for what each of them means.
Here, 1 \leq m \leq 40 and 1 \leq d_i \leq 10^{12} must hold. Also, m and d_i must all be integers.

w_j is a string of length m that represents the way to bring Joint m of the robot arm to point (X_j, Y_j).
The i-th character of w_j should be one of the letters L, R, D and U, representing the mode of Section i.

Sample Input 1

3
-1 0
0 3
2 -1

Sample Output 1

2
1 2
RL
UU
DR

In the given way to bring Joint m of the robot arm to each (X_j, Y_j), the positions of the joints will be as follows:

To (X_1, Y_1) = (-1, 0): First, the position of Joint 0 is (x_0, y_0) = (0, 0). As the mode of Section 1 is R, the position of Joint 1 is (x_1, y_1) = (1, 0). Then, as the mode of Section 2 is L, the position of Joint 2 is (x_2, y_2) = (-1, 0).

To (X_2, Y_2) = (0, 3): (x_0, y_0) = (0, 0), (x_1, y_1) = (0, 1), (x_2, y_2) = (0, 3).

To (X_3, Y_3) = (2, -1): (x_0, y_0) = (0, 0), (x_1, y_1) = (0, -1), (x_2, y_2) = (2, -1).

Sample Input 2

5
0 0
1 0
2 0
3 0
4 0

Sample Output 2

-1

Sample Input 3

2
1 1
1 1

Sample Output 3

2
1 1
RU
UR

There may be duplicated points among (X_j, Y_j).

Sample Input 4

3
-7 -3
7 3
-3 -7

Sample Output 4

5
3 1 4 1 5
LRDUL
RDULR
DULRD

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author __shangu__
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DRobotArms solver = new DRobotArms();
        solver.solve(1, in, out);
        out.close();
    }

    static class DRobotArms {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt(), x[] = new int[n], y[] = new int[n];
            for (int i = 0; i < n; ++i) {
                x[i] = in.nextInt();
                y[i] = in.nextInt();
            }

            int mod2 = Math.abs(x[0]) + Math.abs(y[0]), maxdist = 0;
            mod2 %= 2;
            boolean ok = true;
            for (int i = 1; i < n && ok; ++i) {
                int curdist = Math.abs(x[i]) + Math.abs(y[i]);
                if (curdist % 2 != mod2) ok = false;
                maxdist = Math.max(curdist, maxdist);
            }
            if (!ok) {
                out.println(-1);
                return;
            }
            mod2 = 1 - mod2;

            int dn = 0;
            if (mod2 == 1) maxdist--;
            for (; ((long) 1 << dn) - 1 < maxdist; ++dn) ;
            out.println((mod2 == 1 ? dn + 1 : dn));
            assert (dn > 35);
            if (mod2 == 1) {
                out.print("1 ");
            }
            for (int i = dn - 1; i > 0; --i) out.printf("%d ", (long) 1 << i);
            out.println(1);

            char ans[] = new char[dn + 10];
            for (int i = 0; i < n; ++i) {
                int curx = x[i], cury = y[i], qt = 0;

                if (mod2 == 1) {
                    if (curx > 0) {
                        ans[qt++] = 'R';
                        --curx;
                    } else if (curx < 0) {
                        ans[qt++] = 'L';
                        ++curx;
                    } else if (cury > 0) {
                        ans[qt++] = 'U';
                        --cury;
                    } else {
                        ans[qt++] = 'P';
                        ++cury;
                    }
                }

                for (int j = dn - 1; j >= 0; --j) {
                    if (Math.abs(curx + ((long) 1 << j)) + Math.abs(cury) <= ((long) 1 << j) - 1) {
                        ans[qt++] = 'L';
                        curx += (long) 1 << j;
                    } else if (Math.abs(curx - ((long) 1 << j)) + Math.abs(cury) <= ((long) 1 << j) - 1) {
                        ans[qt++] = 'R';
                        curx -= (long) 1 << j;
                    } else if (Math.abs(curx) + Math.abs(cury + ((long) 1 << j)) <= ((long) 1 << j) - 1) {
                        ans[qt++] = 'D';
                        cury += (long) 1 << j;
                    } else {
                        ans[qt++] = 'U';
                        cury -= (long) 1 << j;
                    }
                }
                for (int j = 0; j < qt; ++j) out.print(ans[j]);
                out.println();
            }
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}