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 <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#define LL long long
using namespace std;
const int maxn = 1e3 + 10;

struct node
{
	LL x, y;
}q[maxn];
int n;
LL b[40];
char ANS[40];
int main()
{
	scanf("%d", &n);
	int c = 0;
	for(int i = 1; i <= n; i ++)
	{
		scanf("%lld %lld", &q[i].x, &q[i].y);
		if((q[i].x + q[i].y) % 2 == 0)
			c --;
		else
			c ++;
	}
	if(abs(c) != n)
	{
		cout << -1;
		return 0;
	}
	int cnt = 32;
	if(c < 0)
		b[33] = 1, cnt ++;
	b[1] = 1;
	for(LL i = 2, temp = 1; i <= 32; i ++)
	{
		temp *= 2;
		b[i] = temp;
	}
	cout << cnt << endl;
	for(int i = 1; i <= cnt; i ++)
		cout << b[i] << " ";
	cout << endl;
	for(int i = 1; i <= n; i ++)
	{
		if(c < 0)
			ANS[cnt] = 'R', q[i].x --;
		for(int j = 32; j >= 1; j --)
		{
			if(abs(q[i].x) >= abs(q[i].y))
			{
				if(q[i].x > 0)
					ANS[j] = 'R', q[i].x -= b[j];
				else
					ANS[j] = 'L', q[i].x += b[j];
			}
			else
			{
				if(q[i].y > 0)
					ANS[j] = 'U', q[i].y -= b[j];
				else
					ANS[j] = 'D', q[i].y += b[j];
			}
		}
		for(int j = 1; j <= cnt; j ++)
			cout << ANS[j];
		cout << endl;
	}
	return 0;
}
