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

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.

Examples

Input

3
-1 0
0 3
2 -1


Output

2
1 2
RL
UU
DR


Input

5
0 0
1 0
2 0
3 0
4 0


Output

-1


Input

2
1 1
1 1


Output

2
1 1
RU
UR


Input

3
-7 -3
7 3
-3 -7


Output

5
3 1 4 1 5
LRDUL
RDULR
DULRD
#include <bits/stdc++.h>

int main()
{
	int N;
	scanf("%d", &N);
	std::vector<int> X(N), Y(N);
	for (int i{}; i < N; i++)
		scanf("%d%d", &X[i], &Y[i]);
	for (int i{}; i < N; i++)
		if (std::abs(X.front() + Y.front()) % 2 != std::abs(X[i] + Y[i]) % 2)
		{
			puts("-1");
			return 0;
		}

	std::vector<int64_t> sum(N), diff(N);
	for (int i{}; i < N; i++)
	{
		sum[i] = X[i] + Y[i];
		diff[i] = X[i] - Y[i];
	}

	std::vector<int64_t> length(32);
	for (int i{}; i < 32; i++)
		length[i] = 1ll << (31 - i);
	if (sum.front() % 2 == 0) length.push_back(1);
	printf("%d\n", length.size());
	for (int i{}; i < (int)length.size(); i++)
		printf("%lld%c", length[i], i + 1 == (int)length.size()? '\n': ' ');

	const char dir[]{"RUDL"};

	for (int i{}; i < N; i++)
	{
		std::vector<int> ans(length.size());
		int64_t place{};
		for (int j{}; j < 32; j++)
			if (place >= sum[i])
			{
				place -= length[j];
				ans[j] += 2;
			}
			else
				place += length[j];
		if (ans.size() % 2 && place > sum[i])
			ans.back() += 2;
		
		place = 0;
		for (int j{}; j < 32; j++)
			if (place >= diff[i])
			{
				place -= length[j];
				ans[j]++;
			}
			else
				place += length[j];
		if (ans.size() % 2 && place > diff[i])
			ans.back()++;
		for (auto& e: ans) putchar(dir[e]);
		putchar('\n');
	}

	return 0;
}