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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  
  int odd = 0, even = 0;
  vector<ll> x(n), y(n);
  for (int i = 0; i < n; i++) {
    cin >> x.at(i) >> y.at(i);
    if ((abs(x.at(i)) + abs(y.at(i))) % 2 == 1) odd++;
    else even++;
  }
  
  if (odd > 0 && even > 0) {
    cout << "-1\n";
    return 0;
  }
  
  vector<ll> arms;
  for (ll i = 35; i >= 0; i--) arms.emplace_back(1LL << i);
  if (even > 0) arms.emplace_back(1);
  
  cout << arms.size() << '\n';
  for (int i = 0; i < arms.size(); i++) {
    if (i) cout << " ";
    cout << arms.at(i);
  }
  cout << '\n';
  
  for (int i = 0; i < n; i++) {
    ll gu = x.at(i) + y.at(i), gv = x.at(i) - y.at(i);
    ll nowx = 0, nowy = 0;
    for (int j = 0; j < arms.size(); j++) {
      ll nowu = nowx + nowy, nowv = nowx - nowy;
      if (nowu < gu && nowv < gv) {
        cout << 'R';
        nowx += arms.at(j);
      }
      else if (nowu > gu && nowv > gv) {
        cout << 'L';
        nowx -= arms.at(j);
      }
      else if (nowu < gu && nowv > gv) {
        cout << 'U';
        nowy += arms.at(j);
      }
      else {
        cout << 'D';
        nowy -= arms.at(j);
      }
    }
    cout << '\n';
  }
}