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<cstdio>
#include<cstring>
#include<algorithm>
#define MN 1005
#define ML 33 
using namespace std;
inline int in(){
    int x=0;bool f=0;char c;
    for (;(c=getchar())<'0'||c>'9';f=c=='-');
    for (x=c-'0';(c=getchar())>='0'&&c<='9';x=(x<<3)+(x<<1)+c-'0');
    return f?-x:x;
}
int x[MN],y[MN],d[ML];
int n,m;
bool par[2];
int main()
{
	n=in();
	for (int i=1;i<=n;++i)
	x[i]=in(),y[i]=in(),par[(x[i]^y[i])&1]=1;
	if (par[0]&&par[1]){printf("-1");return 0;}
	m=31;printf("%d\n",m+par[0]);
	for (int i=30;i>=0;--i)
	d[30-i]=(1<<i),printf("%d ",d[30-i]);
	if (par[0]) d[31]=1,printf("1");puts("");
	for (int i=1;i<=n;++i){
		x[i]+=par[0];
		for (int j=0;j<m;++j){
			if (abs(x[i])>abs(y[i])){
				if (x[i]>0) putchar('R'),x[i]-=d[j];
				else putchar('L'),x[i]+=d[j];
			}else{
				if (y[i]>0) putchar('U'),y[i]-=d[j];
				else putchar('D'),y[i]+=d[j];
			}
		}if (par[0]) putchar('L');puts("");
	}return 0;
}