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>
using namespace std;

int n, m, ct, val[40];
bool s;
char ans[40000];
long long get_dis(pair <int, int> x, pair <int, int> y)
{
	return (long long) abs(x.first - y.first) + abs(x.second - y.second);
}
int main()
{
	m = 32;
	val[1] = 1;
	for (int i = 2; i <= m; i++)
		val[i] = 1 << (i - 2);
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		pair <int, int> now, goal, to;
		scanf("%d%d", &goal.first, &goal.second);
		if (i == 1) s = (goal.first + goal.second) & 1;
		else if (s != ((goal.first + goal.second) & 1))
		{
			puts("-1");
			return 0;
		}
		for (int j = m; j >= (s ? 2 : 1); j--)
		{
			char ch;
			long long mini = 0x3f3f3f3f3f3f3f3fLL, tmp;
			if (mini > (tmp = get_dis(goal, make_pair(now.first, now.second + val[j])))) ch = 'U', mini = tmp, to = make_pair(now.first, now.second + val[j]);
			if (mini > (tmp = get_dis(goal, make_pair(now.first, now.second - val[j])))) ch = 'D', mini = tmp, to = make_pair(now.first, now.second - val[j]);
			if (mini > (tmp = get_dis(goal, make_pair(now.first + val[j], now.second)))) ch = 'R', mini = tmp, to = make_pair(now.first + val[j], now.second);
			if (mini > (tmp = get_dis(goal, make_pair(now.first - val[j], now.second)))) ch = 'L', mini = tmp, to = make_pair(now.first - val[j], now.second);
			now = to;
			ans[++ct] = ch;
		}
		ans[++ct] = '\n';
	}
	printf("%d\n", m - s);
	for (int i = m; i >= (s ? 2 : 1); i--)
		printf("%d ", val[i]);
	puts("");
	for (int i = 1; i <= ct; i++)
		putchar(ans[i]);
	return 0;
}
