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<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};
const char c[]={'R','L','U','D'};
int n;
long long x[1010],y[1010];
long long tx[1010],ty[1010];
int m;
long long l[51];
long long dist(long long a,long long b,long long c,long long d)
{
	return abs(a-c)+abs(b-d);
}
void getl()
{
	long long mx=0;
	for(int i=1;i<=n;i++)
		mx=max(mx,tx[i]+ty[i]);
	while(mx>0)
	{
		l[++m]=(mx+1)/2;
		for(int i=1;i<=n;i++)
			tx[i]-=l[m];
		mx=0;
		for(int i=1;i<=n;i++)
		{
			tx[i]=abs(tx[i]);
			ty[i]=abs(ty[i]);
			if(tx[i]<ty[i])swap(tx[i],ty[i]);
			mx=max(mx,tx[i]+ty[i]);
		}
	}
	return ;
}
char ans[51];
void work(long long a,long long b,int now)
{
	if(now>m)return ;
	long long mn=2e9;
	for(int i=0;i<4;i++)
		mn=min(mn,dist(dx[i]*l[now],dy[i]*l[now],a,b));
	for(int i=0;i<4;i++)
		if(dist(dx[i]*l[now],dy[i]*l[now],a,b)==mn)
		{
			ans[now]=c[i];
			work(a-dx[i]*l[now],b-dy[i]*l[now],now+1);
			break;
		}
	return ;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld %lld",&x[i],&y[i]);
		tx[i]=abs(x[i]);
		ty[i]=abs(y[i]);
		if(tx[i]<ty[i])swap(tx[i],ty[i]);
	}
	for(int i=1;i<=n;i++)
		if(((x[i]+y[i])&1)!=((x[1]+y[1])&1)){
			printf("-1\n");
			return 0;
		}
	getl();
	printf("%d\n",m);
	for(int i=1;i<=m;i++)
		printf("%lld ",l[i]);
	printf("\n");
	for(int i=1;i<=n;i++)
	{
		work(x[i],y[i],1);
		printf("%s\n",ans+1);
	}
	return 0;
}