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<string>
#include<cstring>
#include<utility>
#include<cmath>
#include<map>
#include<queue>
#include<set>
#include<algorithm>
#include<vector>
#include<iostream>
#define ll long long
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define inf 0x7fffffff
using namespace std;
pii a[100010];
int main()
{
	int i,j,k,n,m;
	ll x,y;
	scanf("%d",&n);
	x=0;
	for(i=0;i<n;i++)
	{
		scanf("%d%d",&a[i].fi,&a[i].se);
		x=max(x,1ll*abs(a[i].se)+abs(a[i].fi));
		if(i>0&&(((1ll*abs(a[i].se)+abs(a[i].fi))&1)!=((1ll*abs(a[i-1].se)+abs(a[i-1].fi))&1)))
		{
			printf("-1");
			return 0;
		}
	}
	for(i=0;i<=40;i++)
	{
		if((1ll<<(i))-1>=x)
		{
			break;
		}
	}
	m=i;
	printf("%d\n",x%2==1?i:i+1);
	for(i=0;i<m;i++)
	{
		printf("%d ",1<<i);
	}
	if(x%2==0)
	{
		printf("1");
	}
	printf("\n");
	string s;
	for(i=0;i<n;i++)
	{
		j=0;
		s="";
		if(x%2==0)
		{
			a[i].fi++;
		}
		for(j=0;j<m;j++)
		{
			if(abs(a[i].fi)>abs(a[i].se))
			{
				if(a[i].fi>0)
				{
					s+="R";
					a[i].fi-=(1<<(m-j-1));
				}
				else
				{
					s+="L";
					a[i].fi+=(1<<(m-j-1));
				}
			}
			else
			{
				if(a[i].se>0)
				{
					s+="U";
					a[i].se-=(1<<(m-j-1));
				}
				else
				{
					s+="D";
					a[i].se+=(1<<(m-j-1));
				}
			}
		}
		reverse(s.begin(),s.end());
		cout<<s<<"L "[x%2];
		printf("\n");
	}
}