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>
using namespace std;
const int N=1005;
int dx[]={0,0,1,-1},dy[]={1,-1,0,0};
long long x[N],y[N];
char ch[]={'U','D','R','L'};
int main() 
{
	ios::sync_with_stdio(false); 
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>x[i]>>y[i];
	int m2=0;
	if((x[0]+y[0])%2==0) 
	{
		m2=1;
		for(int i=0;i<n;i++)
			x[i]--;
	}
	for(int i=0;i<n;i++) 
	{
		if((x[i]+y[i])%2==0) 
		{
			cout<<-1<<endl;
			return 0;
		}
	}
	vector<long long>v;
	for(int i=0;i<=31;i++)
		v.push_back(1ll<<i);
	if(m2)
		v.push_back(1);
	reverse(v.begin(),v.end());
	cout<<v.size()<<endl;
	for(auto i=0;i<v.size();i++)
		cout<<v[i]<<" ";
	cout<<endl;
	for(int i=0;i<n;i++) 
	{
		string s;
		if(m2)
			s+="R";
		for(int j=31;j>=0;j--) 
		{
			int tmp=-1;
			long long now=LLONG_MAX;
			for(int k=0;k<4;k++) 
			{
				long long cur=abs(x[i]-dx[k]*(1ll<<j))+abs(y[i]-dy[k]*(1ll<<j));
				if(cur<now)
				{
					now=cur;
					tmp=k;
				}
			}
			x[i]-=dx[tmp]*(1ll<<j);
			y[i]-=dy[tmp]*(1ll<<j);
			s+=ch[tmp];
		}
		cout<<s<<endl;
	}
	return 0;
}
//