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 <bits/stdc++.h>
#define debug(x) cerr << #x << " " << (x) << endl
using namespace std;

const int N = 1001, M = 35;

int n, x[N], y[N];

long long my_abs (long long x) {
	return x > 0 ? x : -x;
}
bool check () {
	for (int i = 1; i < n; i++) {
		if (((1ll << M) + x[i] + y[i]) % 2 != ((1ll << M) + x[n] + y[n]) % 2) return false;
	}
	return true;
}

int main () {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);

	if (!check()) printf("-1\n");
	else {
		if ((x[n] + y[n]) % 2 == 0) printf("%d\n1 ", M + 1);
		else printf("%d\n", M);
		for (int i = 0; i < M; i++) printf("%lld ", 1ll << i);
		printf("\n");
		for (int i = 1; i <= n; i++) {
			string ans;
			if ((x[i] + y[i]) % 2 == 0) x[i]--;

			for (int j = M - 1; j >= 0; j--) {
				long long len = 1ll << j;
				if (my_abs(x[i] - len) + my_abs(y[i]) < len) {
					ans = 'R' + ans;
					x[i] -= len;
				}
				else if (my_abs(x[i] + len) + my_abs(y[i]) < len) {
					ans = 'L' + ans;
					x[i] += len;
				}
				else if (my_abs(x[i]) + my_abs(y[i] - len) < len) {
					ans = 'U' + ans;
					y[i] -= len;
				}
				else if (my_abs(x[i]) + my_abs(y[i] + len) < len) {
					ans = 'D' + ans;
					y[i] += len;
				}
			}

			if ((x[i] + y[i]) % 2 == 0) ans = 'R' + ans;
			cout << ans << endl;

		}
	}
	return 0;
}