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: Wrong Answer)

#include <stdio.h>
#include <math.h>

int main(void)
{
	int n, mod, d1[31], d2[31];
	long x[1000], y[1000], k, u, v;
	char ans[31];
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%ld%ld", &x[i], &y[i]);
	}
	k=2000000000;
	mod=((x[0]+y[0])+k+2)%2;
	for (int i = 0; i < n; ++i)
	{
		if (((x[i]+y[i])+k+2)%2 != mod)
		{
			printf("%d\n", -1);
			return 0;
		}
	}

	if (mod==0)
	{
		for (int i = 0; i < n; ++i)
		{
			x[i]-=1;
		}
		printf("%d ", 1);
		for (int i = 0; i < 31; ++i)
		{
			printf("%ld ", (long)pow(2,i));
		}
		printf("\n");
	}
	else
	{
		for (int i = 0; i < 31; ++i)
		{
			printf("%d ", (int)pow(2,i));
		}
		printf("\n");
	}
	for (int j = 0; j < n; ++j)
	{
		u=(x[j]+y[j]-1)/2+(long)pow(2,30);
		v=(x[j]-y[j]-1)/2+(long)pow(2,30);
		for (int i = 30; i >= 0; i--)
		{
			if (u>=(long)pow(2, i))
			{
				d1[i]=1;
				u-=(long)pow(2, i);
			}
			else
			{
				d1[i]=0;
			}
			if (v>=(long)pow(2, i))
			{
				d2[i]=1;
				v-=(long)pow(2, i);
			}
			else
			{
				d2[i]=0;
			}
			if (d1[i]==1 && d2[i]==1)
			{
				ans[i]='R';
			}
			else if (d1[i]==1 && d2[i]==0)
			{
				ans[i]='U';
			}
			else if (d1[i]==0 && d2[i]==1)
			{
				ans[i]='D';
			}
			else if (d1[i]==0 && d2[i]==0)
			{
				ans[i]='L';
			}
		}
		if (mod==1)
		{
			printf("%s\n", ans);
		}
		else
		{
			printf("R%s\n", ans);
		}
	}
	return 0;
}