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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
using namespace std;

#define ll long long

ll abc (ll x) {
  if (x < 0) {
    return -1 * x;  
  }  
  return x;
}

int main () {
  ll n;
  cin >> n;
  bool ok = false;
  bool fail = false;
  ll x[n], y[n];
  for (ll i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    if (i == 0 && x[i] % 2 == y[i] % 2) {
      ok = true;    
    }  
    if (ok && x[i] % 2 != y[i] % 2) {
      fail = true;    
    }
    if (!ok && x[i] % 2 == y[i] % 2) {
      fail = true;  
    }
  }  
  if (fail) {
    cout << "-1" << endl;  
  }
  else {
  ll m;
  m = 39;
  if (ok) {
    m++;  
  }
  ll two[m];
  for (ll i = 0; i < m; i++) {
    if (i == 0) {
      two[m - i - 1] = 1;  
      cout << m << endl;
    }    
    else {
      if (i == 1 && m == 40) {
        two[m - i - 1] = 1;  
      }
      else {
        two[m - i - 1] = two[m - i] * 2;  
      }
    }
  }
  for (ll i = 0; i < m; i++) {
    if (i != m - 1) {
      cout << two[i] << " ";  
    }  
    else {
      cout << two[i] << endl;  
    }
  }
  for (ll i = 0; i < n; i++) {
    // x, yについて求める.
    for (ll j = 0; j < m; j++) {
      // まずx, yにどっちが大きいかを調べる
      if (abc(x[i]) > abc(y[i])) {
        if (x[i] > 0) {
          x[i] = x[i] - two[j];
          cout << "R";  
        }  
        else {
          x[i] = x[i] + two[j];  
          cout << "L";
        }
      } 
      else {
        if (y[i] > 0) {
          y[i] = y[i] - two[j];
          cout << "U";  
        }    
        else {
          y[i] = y[i] + two[j];
          cout << "D";  
        }
      }         
    }   
    cout << endl;
  }
  }
}