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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#define SZ(x) ((int)(x).size())

typedef long long ll;

int const N = 1e4 + 5;

struct point {
	ll x, y;
}p[N];

int n;

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

bool check() {
	if(n == 1) 
		return 1;
	for(int i = 2; i <= n; ++i)
		if(abs(p[i].x + p[i].y) % 2 != abs(p[1].x + p[1].y) % 2)
			return 0;
	return 1;
}


std::vector<ll> sec;
std::vector<char> ans[N];

int main() {
	// freopen("input.txt", "r", stdin);
	// freopen("output.txt", "w", stdout);
	std::ios::sync_with_stdio(0);
	std::cin >> n;
	for(int i = 1; i <= n; ++i)
		std::cin >> p[i].x >> p[i].y;

	if(::check() == 0) {
		puts("-1");
		return 0;
	}

	for(ll i = 1, bit2 = 1; i <= 34 && bit2 <= ll(2e9); ++i) {
		sec.push_back(bit2);
		bit2 <<= 1;
	}

	if(abs(p[1].x + p[1].y) % 2 == 0) 
		sec.push_back(1);

	for(int i = 1; i <= n; ++i) {
		ans[i].clear();
		for(int j = SZ(sec) - 1; j >= 0; --j) {
			ll const& bit2 = sec[j];
			if(abs(p[i].x) > abs(p[i].y)) {
				if(p[i].x < 0) {
					p[i].x += bit2;
					ans[i].push_back('L');
				}
				else {
					p[i].x -= bit2;
					ans[i].push_back('R');
				}
			} else {
				if(p[i].y < 0) {
					p[i].y += bit2;
					ans[i].push_back('D');
				} 
				else {
					p[i].y -= bit2;
					ans[i].push_back('U');
				}
			}
		}
	}

	std::cout << SZ(sec) << '\n';
	for(auto& x : sec)
		std::cout << x << ' ';
	std::cout << '\n';
	for(int i = 1; i <= n; ++i) {
		for(int j = SZ(ans[i]) - 1; j >= 0; --j)
			std::cout << ans[i][j];
		std::cout << '\n';
	}
	auto chkAns = [] (auto& a, auto& se) -> void {
		point now;
		now.x = now.y = 0;
		int _p = SZ(a) - 1;
		for(auto& d : se) {
			if(a[_p] == 'R') 
				now.x += d;
			if(a[_p] == 'L')
				now.x -= d;
			if(a[_p] == 'U')
				now.y += d;
			if(a[_p] == 'D')
				now.y -= d;
			--_p;
		}
		std::cout << now.x << ' ' << now.y << '\n';
	};
	// std::cout << n << '\n';
	// for(int i = 1; i <= n; ++i)
	//     chkAns(ans[i], sec);
	return 0;
}