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<stdlib.h>

typedef long long int int64;

#define ABS(a) ((a)>0?(a):-(a))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define MAX(a,b) ((a)>(b)?(a):(b))

void run(void){
  int n;
  scanf("%d",&n);
  int *x=(int *)malloc(sizeof(int)*n);
  int *y=(int *)malloc(sizeof(int)*n);
  char *ans=(char *)calloc(n*41,sizeof(char));
  int *dir=(int *)calloc(n,sizeof(int));
  char v[4]={'R','L','U','D'};
  int i;
  for(i=0;i<n;i++) scanf("%d%d",x+i,y+i);
  int d[40];
  int m=0;
  while(m<=40){
    for(i=0;i<n;i++){
      if(ABS(y[i])>ABS(x[i])){
	dir[i]=y[i]>=0?2:3;
      } else {
	dir[i]=x[i]>=0?0:1;
      }
    }
    int min=ABS(x[0])+ABS(y[0]);
    int max=min;
    for(i=1;i<n;i++){
      min=MIN(min,ABS(x[i])+ABS(y[i]));
      max=MAX(max,ABS(x[i])+ABS(y[i]));
    }
    if(max==0) break;
    if((max-min)%2!=0 || (m==40 && max>0)){
      printf("-1\n");
      return;
    }
    if(max==min){
      min=dir[0]>=2?ABS(y[0])-ABS(x[0]):ABS(x[0])-ABS(y[0]);
      max=min;
      for(i=1;i<n;i++){
	min=MIN(min,dir[i]>=2?ABS(y[i])-ABS(x[i]):ABS(x[i])-ABS(y[i]));
	max=MAX(max,dir[i]>=2?ABS(y[i])-ABS(x[i]):ABS(x[i])-ABS(y[i]));
      }
    }
    int dd=(max+min)/2;
    for(i=0;i<n;i++){
      if(dir[i]==0){
	x[i]-=dd;
      } else if(dir[i]==1){
	x[i]+=dd;
      } else if(dir[i]==2){
	y[i]-=dd;
      } else {
	y[i]+=dd;
      }
      ans[41*i+m]=v[dir[i]];
    }
    d[m++]=dd;
  }
  printf("%d\n",m);
  for(i=0;i<m;i++) printf("%d ",d[i]);
  printf("\n");
  for(i=0;i<n;i++) printf("%s\n",ans+i*41);
}

int main(void){
  run();
  return 0;
}