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 abs(int x){
  if(x<0) return -x;
  return x;
}

int main(void){
  int N,f=1,maxsum=0;

  int X[1000],Y[1000],sum[1000];
  scanf("%d",&N);
  int i;
  for(i=0;i<N;i++){
    scanf("%d %d",&X[i],&Y[i]);
    sum[i]=abs(X[i])+abs(Y[i]);
  }
  for(i=0;i<N;i++){
    if(maxsum<sum[i]){
      maxsum=sum[i];
    }
  }
  for(i=0;i<N-1;i++){
    if(maxsum!=sum[i]){
      if((maxsum-sum[i])%2!=0) {
        printf("-1");
        f=0;
        break;
      }
    }
  }
  int j;
  char c;
  if(f){
    printf("%d\n",maxsum);
    for(i=0;i<sum[0];i++){
      printf("1 ");
    }
    printf("\n");
    int cnt,num;
    for(i=0;i<N;i++){
      cnt=0;
      if(X[i]<0 || X[i]>0){
        num=X[i];
        if(X[i]<0){
          c='R';
          num*=-1;
        }
        else c='L';

        for(j=0;j<num;j++){
          printf("%c",c);
          cnt++;
        }

      }
      if(Y[i]<0 || Y[i]>0){
        num=Y[i];
        if(Y[i]<0){
          c='D';
          num*=-1;
        }
        else c='U';

        for(j=0;j<num;j++){
          printf("%c",c);
          cnt++;
        }

      }
      if(cnt<maxsum){
        for(j=0;j<(maxsum-cnt)/2;j++){
          printf("R");
        }
        for(j=0;j<(maxsum-cnt)/2;j++){
          printf("L");
        }
      }
        printf("\n");
    }
  }

  return 0;
}