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;
#define int long long
const int tx[]={0,0,1,-1};
const int ty[]={1,-1,0,0};
int getd(int a,int b,int x,int y){
	return abs(a-x)+abs(b-y);
}
char getc(int &a,int &b,int d,int x,int y){
	int Mn=2e9,op=-1;
	for (int i=0;i<4;i++){
		int xx=a+tx[i]*d,yy=b+ty[i]*d;
		int dis=getd(xx,yy,x,y);
		if (dis<Mn) {Mn=dis; op=i;}
	}
	a=a+tx[op]*d; b=b+ty[op]*d;
	if (op==0) return 'U';
	if (op==1) return 'D';
	if (op==2) return 'R';
	return 'L';
}
int x[1111],y[1111],d[222],cnt[2],m;
signed main(){
	int T; scanf("%lld",&T);
	for (int i=1;i<=T;i++){
		scanf("%lld%lld",&x[i],&y[i]);
		cnt[(x[i]+y[i])%2]++;
	}
	if (cnt[1]&& cnt[0]) return puts("-1"),0;
	for (int i=0;i<=30;i++) d[i+1]=1<<(30-i);
	if (cnt[0]) d[32]=1,m=32; else m=31;
	printf("%lld\n",m);
	for (int i=1;i<=m;i++) printf("%lld ",d[i]); puts("");
	for (int i=1;i<=T;i++){
		int a=0,b=0;
		for (int j=1;j<=m;j++)
			putchar(getc(a,b,d[j],x[i],y[i]));
		puts("");
	}
	return 0;
}
	