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;
typedef long long LL;
const int N=2020;
const LL S=1e10;

void GG(){
	puts("-1");
	exit(0);
}

LL lens[N];

void solve(LL l,LL cx,LL cy,LL x,LL y){
	LL len=lens[l];
	// cerr<<endl<<len<<"\t"<<cx<<'\t'<<cy<<'\t'<<x<<"\t"<<y<<endl;
	if(l==0){
		assert(cx==x);
		assert(cy==y);
		return;
	}
	LL dx=x-cx,dy=y-cy;
	if(dx+dy>0){
		if(dx-dy>0){//R
			putchar('R');
			solve(l-1,cx+len,cy,x,y);
		}else{//U
			putchar('U');
			solve(l-1,cx,cy+len,x,y);
		}
	}else{
		if(dx-dy>0){//D
			putchar('D');
			solve(l-1,cx,cy-len,x,y);
		}else{//L
			putchar('L');
			solve(l-1,cx-len,cy,x,y);
		}
	}
}

LL x[N],y[N];


int main(){
	int n;
	scanf("%d",&n);
	LL mr=0;
	for(int i=0;i<n;i++){
		scanf("%lld%lld",&x[i],&y[i]);
		mr=max(mr,abs(x[i])+abs(y[i]))+10;
	}
	int v=(x[0]+y[0])&1;
	for(int i=1;i<n;i++){
		if(((x[i]+y[i])&1)!=v)GG();
	}
	int m=1;
	lens[m]=1;
	LL rg=0;
	while(rg<mr){
		LL len=rg+1;
		rg=len*2-1;
		lens[++m]=len;
	}
	reverse(lens+1,lens+m+1);
	printf("%d\n",v?m+1:m);
	if(v)printf("1 ");
	for(int i=1;i<=m;i++){
		printf("%lld ",lens[i]);
	}
	printf("\n");
	reverse(lens+1,lens+m+1);
	for(int i=0;i<n;i++){
		if(v)printf("R");
		solve(m,0+v,0,x[i],y[i]);
		printf("\n");
	}
	return 0;
}