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<stdio.h>
#include<algorithm>
#include<math.h>
#include<bits/stdc++.h>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<bitset>
#include<unordered_map> 
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include "boost/algorithm/string.hpp"
#define fio ios_base::sync_with_stdio(false)
#define mod 1000000007
#define mod1 mod
#define mod2 100000009
#define li long long int
#define ll li
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC optimize("Ofast")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define readi(x) scanf("%d",&x)
#define  reads(x)  scanf("%s", x)
#define readl(x) scanf("%I64d",&x)
#define rep(i,n) for(i=0;i<n;i++)
#define revp(i,n) for(i=(n-1);i>=0;i--)
#define myrep1(i,a,b) for(i=a;i<=b;i++)
#define myrep2(i,a,b) for(i=b;i>=a;i--)
#define pb push_back
#define mp make_pair
#define fi first
#define sec second
#define MAXN 1000000000000000000
#define MINN -1000000000000000000
#define INTMAX 1000001000
#define pii pair<ll,ll> 
#define pdd pair<double,double>
#define pic pair<int,char>
#define N 401030
#define lgn 20
#define ddouble long double
#define minus minu
#define PI 3.1415926535
#define lgn 20
 
 
// #define INTMAX 200000000 
 
// using namespace boost;
// #define si short int
 
using namespace std;
using namespace __gnu_pbds;             
typedef priority_queue<pii, vector<pii> ,greater < pii > > min_pq;
// typedef priority_queue<pii, vector<pii> , greater<pii >  > min_pq;
ll toint(const string &s) {stringstream ss; ss << s; ll x; ss >> x; return x;}
string tostring ( ll number ){stringstream ss; ss<< number; return ss.str();}
 
// using namespace __gnu_pbds;         
// typedef priority_queue<pair<ll,char> , vector<pair<ll , char> > > max_pq;
// typedef priority_queue<pii , vector<pii > ,greater<pii > > min_pq;
typedef tree<pii , null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> OST;
// typedef priority_queue<pair < ll , pair < pii , ll > >  , vector<pair < ll , pair < pii , ll > >  > ,greater<pair < ll , pair < pii , ll > >  > > min_pq;
#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cout << name << " : " << arg1 << std::endl;
        //use cerr if u want to display at the bottom
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
#else
#define trace(...)
#endif
#define nn 2000005
ll x[ N ], y[ N ];
pii r[ N ], c[ N ];
string s;
ll n;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    // #ifndef ONLINE_JUDGE
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    // #endif
    cin >> n;
    ll odd = 0,even = 0;
    for ( ll i = 1; i <= n; i ++)
    {
        cin >> x[i] >> y[i];
        if ( (abs(x[i])+abs(y[i]))%2 == 0 )
            even ++;
        else
            odd ++;
    }
    for ( ll i = 0; i < 100000; i ++)
    {
        if ( i %2 == 0 )
            s.pb('R');
        else
            s.pb('L');
    }
    for ( ll i = 1; i <= n; i ++)
    {
        if ( x[i] == 0 )
            r[i]=mp(1LL,1LL);
        else
            r[i]=mp((abs(x[i])+1)/2,(abs(x[i])+1)/2 );
        if ( y[i] == 0 )
            c[i] = mp(1LL,1LL);
        else
            c[i]=mp((abs(y[i])+1)/2  ,(abs(y[i])+1)/2  );
    }
    if ( even == 0 or odd == 0 )
    {
        if ( odd )
        {
            cout << 4*n+1 << endl;
            for ( ll i = 1; i <= n; i ++)
                cout << r[i].fi <<" " << r[i].sec <<" ";
            for ( ll i = 1; i <= n; i ++)
                cout << c[i].fi <<" " << c[i].sec <<" ";
            cout << "1 " << endl;
            for ( ll i = 1; i <= n; i ++)
            {
                for ( ll j = 1; j <= n; j ++)
                {
                    if ( j != i )
                        cout <<"RL";
                    else
                    {
                        if ( x[i] < 0 )
                            cout <<"LL";
                        else if ( x[i] > 0 )
                            cout <<"RR";
                        else
                            cout <<"RL";
                    }
                }
                for ( ll j = 1; j <= n; j ++)
                {
                    if ( j != i )
                        cout <<"RL";
                    else
                    {
                        if ( y[i] < 0 )
                            cout <<"DD";
                        else if ( y[i] > 0 )
                            cout <<"UU";
                        else
                            cout <<"RL";
                    }
                }
                if ( x[i]%2 )
                {
                    if ( x[i] > 0 )
                        cout <<"L";
                    else
                        cout <<"R";
                }
                else
                {
                    if ( y[i] > 0 )
                        cout <<"D";
                    else
                        cout <<"U";
                }
                cout << endl;
            }
        }
        else
        {
            cout << 4*n << endl;
            for ( ll i = 1; i <= n; i ++)
                cout << r[i].fi <<" " << r[i].sec <<" ";
            for ( ll i = 1; i <= n; i ++)
                cout << c[i].fi <<" " << c[i].sec <<" ";
            cout << endl;
            for ( ll i = 1; i <= n; i ++)
            {
                for ( ll j = 1; j <= n; j ++)
                {
                    if ( j != i )
                        cout <<"RL";
                    else
                    {
                        if ( x[i] < 0 )
                            cout <<"LL";
                        else if ( x[i] > 0 )
                            cout <<"RR";
                        else
                            cout <<"RL";
                    }
                }
                for ( ll j = 1; j <= n; j ++)
                {
                    if ( j != i )
                        cout <<"RL";
                    else
                    {
                        if ( y[i] < 0 )
                            cout <<"DD";
                        else if ( y[i] > 0 )
                            cout <<"UU";
                        else
                            cout <<"RL";
                    }
                }
                cout << endl;
            }
        }
    }
    else
    {
        cout <<"-1";
    }





}