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<set>
#include<map>
#include<ctime>
#include<queue>
#include<cmath>
#include<climits>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
int read(){
    int f=1,x=0;char s=getchar();   
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}  
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    return x*f;
}
#define MAXN 1000
#define INF 0x3f3f3f3f
int n,len,x[MAXN+5],y[MAXN+5],d[35];
void Print(int X,int Y){//贪心策略
	for(int i=1;i<=len;i++){
		if(abs(X)>abs(Y)){
			if(X>0) putchar('R'),X-=d[i];
			else putchar('L'),X+=d[i];
		}
		else{
			if(Y>0) putchar('U'),Y-=d[i];
			else putchar('D'),Y+=d[i];
		}
	}
	puts("");
	return ;
}
int main(){
	bool f[2]={0};
	n=read();
	for(int i=1;i<=n;i++)
		x[i]=read(),y[i]=read(),f[(x[i]+y[i])%2]=1;
	if(f[0]&&f[1]){//判断奇偶有无解
		puts("-1");
		return 0;
	}
	for(int i=30;i+1;i--)
		d[++len]=(1<<i);
	if(f[0]) d[++len]=1;//奇偶处理
	printf("%d\n",len);
	for(int i=1;i<len;i++)
		printf("%d ",d[i]);
	printf("%d\n",d[len]);
	for(int i=1;i<=n;i++)
		Print(x[i],y[i]);
	return 0;
}