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 <string>
#include <vector>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

/*
	tar: even
		d_1 = 1, d_{i} = 2^{i-2} (2<=i<=m)
	find s_i in {+1, -1} s.t. s_1 d_1 + ... + s_m d_m = tar
*/
string solve2(int tar,int m){
	if(tar<0){
		string s=solve2(-tar,m);
		for(char &c:s) c=(c=='+')?'-':'+';
		return s;
	}

	if(m==2){
		if(tar==0) return "+-";
		else       return "++";
	}

	return solve2(tar-(1<<(m-2)),m-1)+"+";
}

/*
	tar: even
		m = 32
		d_1 = 1, d_{i} = 2^{i-2} (2<=i<=32)
	tar: odd
		m = 33
		d_1 = 1, d_{i} = 2^{i-2} (2<=i<=32), d_{33} = 1
	find s_i in {+1, -1} s.t. s_1 d_1 + ... + s_m d_m = tar
*/
string solve(int tar){
	if(tar%2!=0) return solve(tar-1)+"+";
	return solve2(tar,32);
}

int main(){
	int n; cin>>n;
	vector<int> x(n),y(n);
	rep(i,n) cin>>x[i]>>y[i];

	int par=((x[0]+y[0])%2+2)%2;
	rep(i,n) if(((x[i]+y[i])%2+2)%2!=par) {
		cout<<"-1"<<endl;
		return 0;
	}

	vector<int> d(1,1);
	rep(i,31) d.emplace_back(1<<i);
	if(par==1) d.emplace_back(1);

	int m=d.size();
	cout<<m<<endl;
	rep(i,m) cout<<d[i]<<' ';
	cout<<endl;

	rep(i,n){
		int u=x[i]+y[i]; // super clever!!
		int v=x[i]-y[i];

		string s1=solve(u);
		string s2=solve(v);

		string ans;
		rep(j,m){
			if     (s1[j]=='+' && s2[j]=='+') ans+='R';
			else if(s1[j]=='+' && s2[j]=='-') ans+='U';
			else if(s1[j]=='-' && s2[j]=='+') ans+='D';
			else if(s1[j]=='-' && s2[j]=='-') ans+='L';
		}
		cout<<ans<<endl;
	}

	return 0;
}
