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: Accepted)

#include <boost/multiprecision/cpp_int.hpp>
#include <cmath>
#include <iostream>
#include <vector>

using bint = boost::multiprecision::cpp_int;

bint power(const int& p, const int& q)
{
    bint result = 1;
    for (int i = 0; i < q; i++) {
        result *= p;
    }
    return result;
}

// bitを小さい位から32桁vectorに詰める
std::vector<int> getBits(const bint& num)
{
    std::vector<int> result{};
    for (int i = 0; i < 32; i++) {
        result.emplace_back(boost::multiprecision::bit_test(num, i));
    }
    return result;
}

int main()
{
    int N;
    std::vector<int> xs;
    std::vector<int> ys;
    std::cin >> N;
    for (int i = 0; i < N; i++) {
        int x, y;
        std::cin >> x >> y;
        xs.emplace_back(x);
        ys.emplace_back(y);
    }
    bool is_odd;
    if (std::abs(xs.at(0) + ys.at(0)) % 2 == 1) {
        is_odd = true;
    } else {
        is_odd = false;
    }
    for (int i = 0; i < N; i++) {
        if (is_odd and (xs.at(i) + ys.at(i)) % 2 == 0) {
            std::cout << -1 << std::endl;
            return 0;
        } else if (not is_odd and std::abs(xs.at(i) + ys.at(i)) % 2 == 1) {
            std::cout << -1 << std::endl;
            return 0;
        }
    }
    if (not is_odd) {
        std::cout << 32 << std::endl;
        for (int i = 0; i < 31; i++) {
            std::cout << power(2, i) << " ";
        }
        std::cout << 1 << std::endl;
    } else {
        std::cout << 31 << std::endl;
        for (int i = 0; i < 30; i++) {
            std::cout << power(2, i) << " ";
        }
        std::cout << power(2, 30) << std::endl;
    }

    for (int i = 0; i < N; i++) {
        int x = xs.at(i);
        int y = ys.at(i);
        // 偶数なら左に一つずらす
        if (not is_odd) {
            x--;
        }
        bint sum_x_plus_y = (power(2, 31) - 1 + x + y) / 2;
        bint sum_x_minus_y = (power(2, 31) - 1 + x - y) / 2;
        std::vector<int> x_plus_y_bits = getBits(sum_x_plus_y);
        std::vector<int> x_minus_y_bits = getBits(sum_x_minus_y);
        for (int j = 0; j < 31; j++) {
            if (x_plus_y_bits.at(j) == 0 and x_minus_y_bits.at(j) == 0) {
                std::cout << "L";
            } else if (x_plus_y_bits.at(j) == 0 and x_minus_y_bits.at(j) == 1) {
                std::cout << "D";
            } else if (x_plus_y_bits.at(j) == 1 and x_minus_y_bits.at(j) == 0) {
                std::cout << "U";
            } else if (x_plus_y_bits.at(j) == 1 and x_minus_y_bits.at(j) == 1) {
                std::cout << "R";
            } else {
                std::cout << x_plus_y_bits.at(j) << " " << x_minus_y_bits.at(j) << std::endl;
                std::cerr << "error" << std::endl;
            }
        }
        if (not is_odd) {
            std::cout << "R" << std::endl;
        } else {
            std::cout << std::endl;
        }
    }


    return 0;
}