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>
#define LL long long
using namespace std;
const int N=1005;
LL a[N],b[N];
char ans[N];
int xx[]={-1,0,1,0},yy[]={0,1,0,-1},zz[]={'L','U','R','D'};
int n,biao[555];
int ck(LL x,LL y){
	return abs(x+y)&1;
}
LL g(LL x,LL y){
	return abs(x)+abs(y);
}
int main(){
	biao['L']='R',biao['R']='L',biao['U']='D',biao['D']='U';
	bool flag=1;
	cin>>n;
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		if(ck(a[i],b[i])!=ck(a[1],b[1]))flag=0;
	}
	if(!flag)puts("-1"),exit(0);
	if(ck(a[1],b[1])==0){
		for(int i=1;i<=n;i++)a[i]--;
		flag=0;
		printf("33\n1 ");
	}
	else{
		printf("32\n");
	}
	for(int i=0;i<=31;i++)printf("%lld%c",1LL<<i," \n"[i==31]);
	for(int k=1;k<=n;k++){
		if(!flag)putchar('R');
		int cnt=0;
		LL p=a[k],q=b[k],now=0;//
		for(int i=31;i>=0;i--){
			for(int j=0;j<4;j++){
				if(g(a[k]+xx[j]*(1LL<<i),b[k]+yy[j]*(1LL<<i))<=(1LL<<i)-1){
					a[k]=a[k]+xx[j]*(1LL<<i);
					b[k]=b[k]+yy[j]*(1LL<<i);
					ans[++cnt]=zz[j];
					break;
				}
			}
		}
		assert(a[k]==0&&b[k]==0);
		for(int i=cnt;i>=1;i--){
			putchar(biao[ans[i]]);
			++now;//
			if(ans[i]=='L')p-=(1LL<<(now-1));
			if(ans[i]=='R')p+=(1LL<<(now-1));
			if(ans[i]=='U')q+=(1LL<<(now-1));
			if(ans[i]=='D')q-=(1LL<<(now-1));
		}
		assert(p==0&&q==0);
		puts("");
	}
}