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.math.*;
import java.util.*;

class Main {
    private static final Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int n = Integer.parseInt(sc.nextLine());
        long[] xs = new long[n];
        long[] ys = new long[n];
        int odd = 0, even = 0;
        for (int i = 0; i < n; i++) {
            String[] xy = sc.nextLine().split(" +");
            xs[i] = Long.parseLong(xy[0]);
            ys[i] = Long.parseLong(xy[1]);
            if ((xs[i] + ys[i]) % 2 == 0)  {
                even++;
                xs[i]--;
            }
            else odd++;
        }

        if (odd * even != 0) {
            System.out.println(-1);
        }
        else {
            System.out.println(even > 0 ? 32 : 31);
            if (even > 0) System.out.print("1 ");
            for (long bit = 1; bit < (1L << 31); bit <<= 1)
                System.out.printf("%d%c", bit, bit == (1 << 30) ? '\n' : ' ');

            for (int i = 0; i < n; i++) {
                long u = xs[i] + ys[i];
                long v = xs[i] - ys[i];

                long maskU = (u + (1 << 31) - 1) >> 1;
                long maskV = (v + (1 << 31) - 1) >> 1;

                long x = 0, y = 0;
                if (even > 0) {
                    System.out.print("R");
                    x++;
                }
                for (long bit = 1; bit < (1L << 31); bit <<= 1) {
                    if ((maskU & bit) == 0 && (maskV & bit) == 0) {
                        System.out.print("L");
                        x -= bit;
                    }
                    else if ((maskU & bit) != 0 && (maskV & bit) != 0) {
                        System.out.print("R");
                        x += bit;
                    }
                    else if ((maskU & bit) == 0 && (maskV & bit) != 0) {
                        System.out.print("D");
                        y -= bit;
                    }
                    else if ((maskU & bit) != 0 && (maskV & bit) == 0) {
                        System.out.print("U");
                        y += bit;
                    }
                }
                System.out.println();
                //System.out.printf("x=%d y=%d\n", x, y);
            }
        }
        sc.close();
    }
}