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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <set>
#include <list>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;A++)

using namespace std;

int i,j,k,l,m,n;
/////////////////////////////////////////////////////

ll N;
ll X[1010],Y[1010];
vector<ll> nv;

ll Abss(ll a){
    return (a>0?a:-a);
}

void errors(){
    printf("-1\n");
    exit(0);
}

void Lwritter(int* checker,ll* nvs,ll num,int start){
    //printf("N%lld\n",num);
    ll pos=num;
    ll a=num;
    for(int i=start;i<nv.size();i++){
        if(Abss(pos)%(2*nv[i])==nv[i]){
            pos-=nvs[i];
            checker[i]=1;
            if(pos==0)
                return;
        }
    }
}

int main(){
    scanf("%lld",&N);
    fornum(i,0,N){
        scanf("%lld%lld",&X[i],&Y[i]);
    }
    int odd=Abss(X[0]+Y[0])%2;
    ll maxnum=Abss(X[0])>Abss(Y[0])?Abss(X[0]):Abss(Y[0]);
    fornum(i,1,N){
        if(odd!=Abss(X[i]+Y[i])%2){
            errors();
        }
        maxnum=(maxnum<Abss(X[i])?Abss(X[i]):maxnum);
        maxnum=(maxnum<Abss(Y[i])?Abss(Y[i]):maxnum);
    }
    //printf("%lld\n",maxnum);
    ll maxx;
    if(odd){
        nv.push_back(1);
    }
    nv.push_back(1);
    for(maxx=1;maxx<maxnum;maxx*=2){
        nv.push_back(maxx);
    }
    nv.push_back(maxx);
    maxx*=2;

    printf("%d\n",(int)(nv.size()));
    fornum(i,0,nv.size()){
        printf("%lld ",nv[i]);
    }
    

    fornum(i,0,N){
        ll summ=Abss(X[i])+Abss(Y[i])-odd;
        int xc[50];
        ll nvc[50];
        fornum(j,0,50){
            xc[j]=0;
        }
        ll a=(maxx-summ)/2;
        //printf("%lld,%lld,%lld",maxx,summ,a);
        j=1;
        nvc[0]=1;
        if(odd){
            nvc[1]=1;
            j++;
        }
        for(j;j<nv.size();j++){
            if(a%2){
                nvc[j]=-nv[j];
            }else{
                nvc[j]=nv[j];
            }
            a/=2;
        }

        if(Abss(X[i])<Abss(Y[i])){
            Lwritter(xc,nvc,(Abss(X[i])%2?Abss(X[i])-odd:Abss(X[i])),odd);
        }else{
            Lwritter(xc,nvc,(Abss(Y[i])%2?Abss(Y[i])-odd:Abss(Y[i])),odd);
            fornum(j,odd,nv.size()){
                xc[j]=!xc[j];
            }
        }
        
        printf("\n");
        //printf("\nX%lld,Y%lld",X[i],Y[i]);
        j=0;
        if(odd){
            if(Abss(X[i])%2){
                if(X[i]>0){
                    printf("R");
                }else{
                    printf("L");
                }
            }else{
                if(Y[i]>0){
                    printf("U");
                }else{
                    printf("D");
                }
            }
            j++;
        }

        for(j;j<nv.size();j++){
            //printf("\n%lld,%d:",nvc[j],xc[j]);
            if(X[i]*Y[i]==0){
                printf("%c",nvc[j]>0?'U':'D');
                continue;
            }
            if(xc[j]==1){
                if(nvc[j]*X[i]>0){
                    printf("R");
                }else{
                    printf("L");
                }
            }else{
                if(nvc[j]*Y[i]>0){
                    printf("U");
                }else{
                    printf("D");
                }
            }
        }
    }

    return 0;
}