Problem:
Score : 600 points

Problem Statement

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

Partial Score

In the test cases worth 300 points, -10 \leq X_i \leq 10 and -10 \leq Y_i \leq 10 hold.

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.

Sample Input 1

3
-1 0
0 3
2 -1

Sample Output 1

2
1 2
RL
UU
DR

In the given way to bring Joint m of the robot arm to each (X_j, Y_j), the positions of the joints will be as follows:

To (X_1, Y_1) = (-1, 0): First, the position of Joint 0 is (x_0, y_0) = (0, 0). As the mode of Section 1 is R, the position of Joint 1 is (x_1, y_1) = (1, 0). Then, as the mode of Section 2 is L, the position of Joint 2 is (x_2, y_2) = (-1, 0).

To (X_2, Y_2) = (0, 3): (x_0, y_0) = (0, 0), (x_1, y_1) = (0, 1), (x_2, y_2) = (0, 3).

To (X_3, Y_3) = (2, -1): (x_0, y_0) = (0, 0), (x_1, y_1) = (0, -1), (x_2, y_2) = (2, -1).

Sample Input 2

5
0 0
1 0
2 0
3 0
4 0

Sample Output 2

-1

Sample Input 3

2
1 1
1 1

Sample Output 3

2
1 1
RU
UR

There may be duplicated points among (X_j, Y_j).

Sample Input 4

3
-7 -3
7 3
-3 -7

Sample Output 4

5
3 1 4 1 5
LRDUL
RDULR
DULRD

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

typedef char                SINT8;
typedef unsigned char       UINT8;
typedef short               SINT16;
typedef unsigned short      UINT16;
typedef int                 SINT32;
typedef unsigned int        UINT32;
typedef long long           SINT64;
typedef unsigned long long  UINT64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))


SINT32 main() {
    SINT32 m;

    SINT64 MAX = 0;

    SINT64 max2 = 1;

    scanf("%d",&m);

    SINT32 ans = 0;

    SINT32 x[100001] = {0};
    SINT32 y[100001] = {0};

    for (SINT32 i = 0; i < m; i++) {
        scanf("%d", &x[i]);
        scanf("%d", &y[i]);

        if (MAX < (ABS(x[i]) + ABS(y[i]))) {
            MAX = ABS(x[i]) + ABS(y[i]);
        }


        if (ans == 0) {
            if ((ABS(x[i] + y[i]) % 2) == 1) {
                ans = 1;
            } else {
                ans = 2;
            }
        } else if (ans == 1) {
            if ((ABS(x[i] + y[i]) % 2) == 0) {
                ans = -1;
                break;
            }
        } else {
            if ((ABS(x[i] + y[i]) % 2) == 1) {
                ans = -1;
                break;
            }
        }
    }
    if (ans == -1) {
        printf("-1\n");
	    return 0;
    }

    SINT32 count = 0;

    while (max2 < MAX) {
        count++;
        max2 = max2 * 2;
    }
    
    SINT64 ans3 = max2;
    if (ans == 1) {
        printf("%d\n",count+1);
        for (SINT32 i = 0; i < (count + 1); i++) {
            if (i == count ) {
                printf("%lld",1);
            } else {
                printf("%lld ",ans3);
                ans3 = ans3/2;
            }
        }
    } else {
        printf("%d\n",count+2);
        for (SINT32 i = 0; i < (count + 2); i++) {
            if (i == count + 1) {
                printf("%lld",1);
            } else {
                printf("%lld ",ans3);
                ans3 = ans3/2;
            }
        }
    }
    printf("\n");


    if (ans == 1) {
        for ( int i = 0; i < m; i++ ) {
            ans3 = max2;
            for (SINT32 j = 0; j < (count + 1); j++) {
                if (ABS(x[i]) > ABS(y[i])) {
                    if (x[i] >= 0) {
                        x[i] = x[i] - ans3;
                        printf("R");
                    } else {
                        x[i] = x[i] + ans3;
                        printf("L");
                    }
                } else {
                    if (y[i] >= 0) {
                        y[i] = y[i] - ans3;
                        printf("U");
                    } else {
                        y[i] = y[i] + ans3;
                        printf("D");
                    }
                }
                ans3 = ans3/2;
            }
            printf("\n");
        }
    } else {
        for ( int i = 0; i < m; i++ ) {
            ans3 = max2;
            for (SINT32 j = 0; j < (count + 2); j++) {
                if (ABS(x[i]) > ABS(y[i])) {
                    if (x[i] >= 0) {
                        x[i] = x[i] - ans3;
                        printf("R");
                    } else {
                        x[i] = x[i] + ans3;
                        printf("L");
                    }
                } else {
                    if (y[i] >= 0) {
                        y[i] = y[i] - ans3;
                        printf("U");
                    } else {
                        y[i] = y[i] + ans3;
                        printf("D");
                    }
                }
                if (i == count + 1) {
                    ans3 = 1;
                } else {
                    ans3 = ans3/2;
                }
            }
            printf("\n");
        }
    }

	return 0;
}