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<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define maxn 1010
using namespace std;
typedef long long ll;
int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(ch-'0'<0||ch-'0'>9){if(ch=='-') f=-1;ch=getchar();}
    while(ch-'0'>=0&&ch-'0'<=9){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n;
struct P{
    int x,y;
}p[maxn];
int a[maxn],cnt;
char ans[maxn],top;
void work(int x,int y)
{
    top=0;
    for(int i=cnt;i>=1;i--)
    {
        if(abs(x)>abs(y))
        {
            if(x<0)
            {
                x+=a[i];
                ans[++top]='L';
            }
            else{
                x-=a[i];
                ans[++top]='R';
            }
        }
        else{
            if(y<0)
            {
                y+=a[i];
                ans[++top]='D';
            }
            else{
                y-=a[i];
                ans[++top]='U';
            }
        }
    }
    for(int i=top;i>=1;i--)  printf("%c",ans[i]);
    puts("");
}
int main()
{
    n=read();
    for(int i=1;i<=n;i++)  p[i].x=read(),p[i].y=read();
    int com=((p[1].x+p[1].y)&1);
    for(int i=1;i<=n;i++)
    {
        if(((p[i].x+p[i].y)&1)!=com)
        {
            puts("-1");
            return 0;
        }
    }
    int mx=0;
    for(int i=1;i<=n;i++)  mx=max(mx,p[i].x+p[i].y-(!com));
    if(com==0)  a[++cnt]=1;
    for(int i=0;;i++)
    {
        if((1ll<<i)>mx)  break;
        a[++cnt]=(1<<i);
    }
    printf("%d\n",cnt);
    for(int i=1;i<=cnt;i++)  printf("%d ",a[i]);
    puts("");
    for(int i=1;i<=n;i++)  work(p[i].x,p[i].y);
    return 0;
}
