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;
long long n,a[1003],b[1003];
long long Dis(long long x,long long y){
	return fabs(x)+fabs(y);
}
pair<long long,long long>lrud(long long x,long long y){
	long long a=pow(2,30);
	while(a){
		pair<long long,long long>c[4]={make_pair(Dis(x+a,y),0),make_pair(Dis(x-a,y),1),make_pair(Dis(x,y+a),2),make_pair(Dis(x,y-a),3)};
		sort(c,c+4);
		if(c[0].second==0)x+=a,cout<<"L";
		if(c[0].second==1)x-=a,cout<<"R";
		if(c[0].second==2)y+=a,cout<<"D";
		if(c[0].second==3)y-=a,cout<<"U";
		a/=2;
	}
	return make_pair(x,y);
}
int main(){
	cin>>n;
	for(long long i=0;i<n;i++)cin>>a[i]>>b[i];
	for(long long i=1;i<n;i++)if(int(fabs(a[i]+b[i]))%2!=int(fabs(a[i-1]+b[i-1]))%2){cout<<-1<<' ';return 0;}
	if((a[0]+b[0])%2==0){
		cout<<32<<endl;
		long long xx=pow(2,30),cnt=0;
		while(xx!=1)cout<<xx<<' ',xx/=2,cnt++;
		cout<<1<<' '<<1<<endl;
		for(long long i=0;i<n;i++){
			pair<long long,long long>x=lrud(a[i],b[i]);
			if(x.first==1)cout<<"R";
			if(x.first==-1)cout<<"L";
			if(x.second==1)cout<<"U";
			if(x.second==-1)cout<<"D";
			if(x.first==2)cout<<"RR";
			if(x.first==-2)cout<<"LL";
			if(x.second==2)cout<<"UU";
			if(x.second==-2)cout<<"DD";
			if(!x.first&&!x.second)cout<<"LR";
			cout<<endl;
		}
	}else{
		cout<<"31\n";
		long long xx=pow(2,30);
		while(xx!=1)cout<<xx<<' ',xx/=2;
		cout<<"1\n";
		for(long long i=0;i<n;i++)
		{
			pair<long long,long long>x=lrud(a[i],b[i]);
			if(x.first==1)cout<<"R";
			if(x.first==-1)cout<<"L";
			if(x.second==1)cout<<"U";
			if(x.second==-1)cout<<"D";
			cout<<endl;
		}
	}
}