Problem: 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

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.

Examples

Input

3
-1 0
0 3
2 -1


Output

2
1 2
RL
UU
DR


Input

5
0 0
1 0
2 0
3 0
4 0


Output

-1


Input

2
1 1
1 1


Output

2
1 1
RU
UR


Input

3
-7 -3
7 3
-3 -7


Output

5
3 1 4 1 5
LRDUL
RDULR
DULRD

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

    static int N;
    static int[] X;
    static int[] Y;

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        N = sc.nextInt();
        X = new int[N];
        Y = new int[N];
        for (int i = 0; i < N; i++) {
            X[i] = sc.nextInt();
            Y[i] = sc.nextInt();
        }

        Answer answer = solve();
        PrintWriter pw = new PrintWriter(System.out);
        if( answer != null ) {
            pw.println(answer.m);
            StringJoiner j = new StringJoiner(" ");
            for (int i : answer.d) {
                j.add(String.valueOf(i));
            }
            pw.println(j.toString());
            for (String w : answer.w) {
                pw.println(w);
            }
        } else {
            pw.println(-1);
        }
        pw.flush();
    }

    static class Answer {
        int m;
        int[] d;
        String[] w;

        public Answer(int m, int[] d, String[] w) {
            this.m = m;
            this.d = d;
            this.w = w;
        }
    }

    static Answer solve() {
        int max = Math.abs(X[0]) + Math.abs(Y[0]);
        boolean even = max % 2 == 0;
        for (int i = 1; i < N; i++) {
            int xy = Math.abs(X[i]) + Math.abs(Y[i]);
            if( xy % 2 == 0 != even ) {
                return null;
            }
            max = Math.max(xy, max);
        }

        int bit;
        int m;
        int[] d;
        if( even ) {
            bit = Integer.toBinaryString(max-1).length();
            m = bit+1;
            d = new int[m];
            for (int b = bit-1; b >= 0 ; b--) {
                d[bit-b-1] = 1 << b;
            }
            d[m-1] = 1;

        } else {
            bit = Integer.toBinaryString(max).length();
            m = bit;
            d = new int[m];
            for (int b = bit-1; b >= 0 ; b--) {
                d[bit-b-1] = 1 << b;
            }
        }

        String[] w = new String[N];
        for (int i = 0; i < N; i++) {
            int x = X[i];
            int y = Y[i];

            StringBuilder sb = new StringBuilder();
            for (int arm : d) {
                if( Math.abs(x) > Math.abs(y) ) {
                    if( x > 0 ) {
                        x = x - (arm);
                        sb.append("R");

                    } else {
                        x = x + (arm);
                        sb.append("L");
                    }
                } else {
                    if( y > 0 ) {
                        y = y - (arm);
                        sb.append("U");

                    } else {
                        y = y + (arm);
                        sb.append("D");
                    }
                }
            }

            if( x == 0 && y == 0 ) {
                w[i] = sb.toString();
            } else {
                throw new RuntimeException("?????!?!? : (" + x + " " + y + ")");
            }
        }

        return new Answer(m, d, w);
    }

    @SuppressWarnings("unused")
    static class FastScanner {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

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

        String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken("\n");
        }

        long nextLong() {
            return Long.parseLong(next());
        }

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

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }
}