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: Runtime Error)


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <limits.h>
#define inf (INT_MAX-1)
#define INF 9223372036854775807
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r);
#define chsort(s,n) qsort(s,n,sizeof(char),cmp)
#define chsort_r(s,n) qsort(s,n,sizeof(char),char_cmp_r);
#define TYPE long long
#define MEMSET(a) memset(a,0,sizeof(a))
const int mod=(int)1e09+7;

int in(void){
    int i;scanf("%d",&i);
    return i;
}
double din(void){
    double i;scanf("%lf",&i);
    return i;
}
void chin(char s[]){
    scanf("%s",s);
}
void print(int a){
    printf("%d\n",a);
}
void llprint(long long a){
    printf("%lld\n",a);
}
void dprint(double a){
    printf("%.10f\n",a);
}
void print2(int a,int b){
    printf("%d %d\n",a,b);
}
double dmax(double a,double b){
    return a>b?a:b;
}

int char_cmp(const void *a,const void *b){
    return strcmp((char *)a,(char *)b);
}
int char_cmp_r(const void *a,const void *b){
    return strcmp((char *)b,(char *)a);
}
void swap(int *a,int *b){
    int t=*a;
    *a=*b;
    *b=t;
}

int intpow(int a, int b){
    int i, c=1;
    rep(i, b){
        c *= a;
    }
    return c;
}

int main(void){
    int i, j, n, m=31, ans=-1, eval[2], f=0;             //0<m<41
    n = in();
    int x[n], y[n], d[31];                               //0<d_i<10^12+1
    rep(i,n){
        x[i] = in();                                     //x+y=u, x-y=v
        y[i] = in();                                     //u_i+1=x_i+y_i±d_i, v_i+1=x_i-y_i±d_i
        eval[i%2] = (x[i]+y[i])%2;
        if(i>0&&eval[0]!=eval[1]) f = 1;
    }
    if(f==0){
        if((x[0]+y[0])%2==0) m++;
        print(m);
        rep(i, 31){
            if((x[0]+y[0])%2!=0||i>0){
                d[i] = intpow(2, i);
                printf("%d ", d[i]);
            }
            else d[i]=0;
        }
        printf("\n");
        int u[32], v[32];
    
        rep(i, n){
            u[31]=x[i]+y[i]+intpow(2, 31)-1;              //u[30]=2*(d_a+d_b+...d_c)-2^31+1
            v[31]=x[i]-y[i]+intpow(2, 31)-1;              //v[30]=2*(d_a'+d_b'+...d_c')-2^31+1
            int a[31], b[31];                             //a[j]:d_jの正負, b[j]:d_jの正負
            rep(j, 31){
                if(abs(u[31-j]-d[30-j])<=d[30-j]){
                    a[30-j]=1;
                }
                else a[30-j]=-1;
                if(abs(v[i]-d[30-j])<=d[30-j]){
                    b[30-j]=1;
                }
                else b[30-j]=-1;
                u[30-j] = u[31-j]-d[30-j]*a[30-j];
                v[30-j] = v[31-j]-d[30-j]*b[30-j];
            }
            int x_1[32], y_1[32];
            for(j=1;j<32;j++){
                u[j]=u[j-1]+a[j-1]*d[j-1];
                v[j]=v[j-1]+a[j-1]+d[j];
                x_1[j]=u[j]+v[j];
                y_1[j]=u[j]-v[j];
                if(x_1[j]-x_1[j-1]<0) printf("L");
                else if(x_1[j]-x_1[j-1]>0) printf("R");
                else if(y_1[j]-y_1[j-1]<0) printf("D");
                else if(y_1[j]-y_1[j-1]>0) printf("U");
            }
            printf("\n");
        }
    }
    if(f==1) print(ans);
    return 0;
}