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 <stdio.h>
#include <string.h>
#include <stdlib.h>
//#include <math.h>
#include <algorithm>
//#include <vector> 

int main(int argc, char* argv[])
{
	long n;
    scanf("%ld", &n);

    long flag0=0, flag1=0;
    long x[1000] = {0};
    long y[1000] = {0};
	long i;
	for(i=0; i<n; i++) {
        scanf("%ld %ld", &(x[i]), &(y[i]));
        if(abs(x[i]+y[i])%2==0) {
            flag0=1;
        }
        else {
            flag1=1;
        }
	}
    if(flag0 && flag1) {
        printf("-1\n");
        return 0;
    }

    long d[32]={0};
    d[30] = d[31] = 1;
    for(i=29; i>=0; i--) {
        d[i] = d[i+1] * 2;
    }
    long m = (flag1)? 31: 32;

    printf("%d\n", m);
    for(i=0; i<m; i++) {
        printf("%d", d[i]);
        if(i<m-1) {
            printf(" ");
        }
    }
    printf("\n");

    for(i=0; i<n; i++) {
        long j;
        for(j=0; j<m; j++) {
            if( abs(x[i]) >= abs(y[i]) ) {
                if(x[i]>=0) {
                    printf("R");
                    x[i] -= d[j];
                }
                else{
                    printf("L");
                    x[i] += d[j];
                }
            }
            else {
                if(y[i]>0) {
                    printf("U");
                    y[i] -= d[j];
                }
                else{
                    printf("D");
                    y[i] += d[j];
                }
            }
        }
        printf("\n");
        //printf("%d %d\n", x[i], y[i]);
    }

    return 0;
}
