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
//ios::sync_with_stdio(false);
#include<bits/stdc++.h>
#define LL long long
#define F(x,y,z) for(int x=y;x<=z;++x)
#define D(x,y,z) for(int x=y;x>=z;--x)
#define ckmax(a,b) (a)=max(a,b)
#define ckmin(a,b) (a)=min(a,b)
using namespace std;
const int N=1010;
LL R(){
    LL ans=0,f=1;char c=getchar();
    for(;c<'0'||c>'9';c=getchar()) if (c=='-') f=-1;
    for(;c>='0'&&c<='9';c=getchar()) ans=ans*10+c-'0';
    return ans*f;
}
LL n,fg[2],tp;
LL a[N],b[N],p[N];
char an[N];
void Work(LL x,LL y){
	F(i,1,tp){
		if(abs(x)>abs(y)){
			if(x>0){
				an[i]='R';
				x-=p[i];
			}
			else{
				an[i]='L';
				x+=p[i];
			}
		}
		else{
			if(y>0){
				an[i]='U';
				y-=p[i];
			}
			else{
				an[i]='D';
				y+=p[i];
			}
		}
	}
	F(i,1,tp)cout<<an[i];
	cout<<'\n';
}
int main(){
	//freopen("1.in","r",stdin);freopen("1.out","w",stdout);
    n=R();
    F(i,1,n){
    	a[i]=R(),b[i]=R();
    	fg[(a[i]+b[i])&1]=1;
	}
	if(fg[0]&&fg[1]){cout<<-1;return 0;}
	F(i,0,30)p[31-i]=(1<<i);
	tp=31;
	if(fg[0])p[++tp]=1;
	cout<<tp<<'\n';
	F(i,1,tp)cout<<p[i]<<' ';
	cout<<'\n';
	F(i,1,n)Work(a[i],b[i]);
	return 0;
}