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>
#define MAXN 1010
using namespace std;

int n,x[MAXN],y[MAXN],tx,ty,p;
vector <int> v;

inline int abs(int a){return a<0?-a:a;}

inline int read()
{
	int x=0;char ch=getchar();bool f=false;
	while (ch<'0'||ch>'9'){if (ch=='-') f=true;ch=getchar();}
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return f?-x:x;
}

int main()
{
	n=read();
	x[1]=read();y[1]=read();p=(abs(x[1])+abs(y[1]))%2;
	for (int i=2;i<=n;i++)
	{
		x[i]=read();y[i]=read();
		if ((abs(x[i])+abs(y[i]))%2!=p)
		{
			puts("-1");
			return 0;
		}
	}
	for (int i=30;~i;i--) v.push_back(1<<i);
	if (!p) v.push_back(1);
	printf("%d\n",v.size());
	for (int i=0;i<v.size();i++) printf("%d%c",v[i],i==v.size()-1?'\n':' ');
	for (int i=1;i<=n;i++)
	{
		tx=x[i];ty=y[i];
		for (int j=0;j<v.size();j++)
		{
			if (abs(tx)>abs(ty))
			{
				if (tx>0){tx-=v[j];putchar('R');}
				else{tx+=v[j];putchar('L');}
			}
			else
			{
				if (ty>0){ty-=v[j];putchar('U');}
				else{ty+=v[j];putchar('D');}
			}
		}
		putchar('\n');
	}
	return 0;
}