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 C (GCC 5.4.1) (judge verdict: Output Limit Exceeded)

#include<stdio.h>
#define abs(x) (x>=0) ? (x) : (-(x))

void writexy(int px1,int px2);

int main()
{
    int n,i,j,k,a,b,odd,flag=1,man;
	int px[1010],py[1010];
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		scanf("%d %d",&a,&b);
		px[i]=a;
		py[i]=b;
		if(i==0)odd=(abs((a+b)))%2;
		else if(odd!=(abs((a+b)))%2)flag=0;
		
	}
	if(!flag)
	{
		printf("-1");
		return 0;
	}
	else
	{
		if(odd==1)
		{
			printf("19\n");
			printf("1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n");//19
			for(j=0;j<n;j++)
			{
				man=((abs(px[j]))+(abs(py[j])));
				for(k=0;k<((19-man)/2);k++)printf("UD");
				writexy(px[j],py[j]);
				printf("\n");
			}
		}
		else if(odd==0)
		{
			printf("20\n");
			printf("1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n");//20
			for(j=0;j<n;j++)
			{
				man=((abs(px[j]))+(abs(py[j])));
				for(k=0;k<((20-man)/2);k++)printf("UD");
				writexy(px[j],py[j]);
				printf("\n");
			}
		}
	}
}

void writexy(int px1,int px2)
{
	int l;
	for(l=0;l<(abs(px1));l++)
	{
		if(px1>0)printf("R");
		else printf("L");
	}
	for(l=0;l<(abs(px2));l++)
	{
		if(px2>0)printf("U");
		else printf("D");
	}
}