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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REPR(i,n) for(int i=n; i>-1; --i)
#define ALL(a) (a).begin(),(a).end()
#define FILL(a,n,x); REP(i,(n)){ (a)[i]=(x); }
#define CINA(a,n); REP(i,(n)){ cin >> (a)[i]; }
#define FILL2(a,n,m,x); REP(i,(n)){ REP(j,(m)){(a)[i][j]=(x);} }
#define CINA2(a,n,m); REP(i,(n)){ REP(j,(m)){cin >> (a)[i][j];} }
#define Liny "Yes\n"
#define Linn "No\n"
#define LINY "YES\n"
#define LINN "NO\n"

//cout << setfill('0') << right << setw(4) << 12; // "0012"
int keta(ll x){ if(x<10){return 1;} else{return keta(x/10) + 1;}}
int keta_wa(ll x){ if(x<10){return x;} else{return keta_wa(x/10) + x%10;} }
int ctoi(char c){ return ( (c>='0' && c<='9')? c - '0': 0 );}
int __stoi(string s){ return atoi(s.c_str()); }
ll sum(ll a[],ll N){ return accumulate(a,a+N,0LL);}
ll gcd(ll a,ll b){if(a<b)swap(a,b); return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){if(a<b){swap(a,b);} return a/gcd(a,b)*b;}
template<class T> void chmax(T& a, T b){ if(a<b){a=b;} }
template<class T> void chmin(T& a, T b){ if(a>b){a=b;} }
template<class T> bool isIn(T a,vector<T> v){ for(T x:v){ if(a==x){return true;}} return false;}
string strReplace(string s,string target, string replacement){
    if (!target.empty()) {
      std::string::size_type pos = 0;
      while ((pos = s.find(target, pos)) != std::string::npos) {
        s.replace(pos, target.length(), replacement);
        pos += replacement.length();
      }
    }
    return s;
}
const ll MOD = 1e9+7;
#define pii pair<int,int>
#define pll pair<ll,ll>
#define MP make_pair

bool DEBUG = 0;

template<class T> void dprint(T s){
  if(DEBUG){ cout << s << "\n"; }
}


int main(int argc,char* argv[]){
  if(argc>1 && argv[1][0]=='D'){ DEBUG=1;}
  dprint("start");

  int N; cin>>N;
  ll x[N],y[N],u[N],v[N];
  REP(i,N){
    cin >> x[i] >> y[i];
  }
  
  int odd = abs(x[0]+y[0])%2;
  REP(i,N){
    if( abs(x[i]+y[i])%2!=odd ){
      cout << -1 << "\n"; return 0;
    }
    u[i] = x[i]+y[i];
    v[i] = x[i]-y[i];
  }
  vector<ll> d;
  REPR(i,30){
    d.push_back(pow(2,i));
  }
  if(!odd){ d.push_back(1LL); }
  int m = d.size();
  cout << m << "\n";
  REP(i,m){
    if(i<m-1){
      cout << d[i] << " ";
    }else{
      cout << d[i] << "\n";
    }
  }

  REP(i,N){
    bool ru,rd;
    ll us = 0LL, vs = 0LL;
    REP(idx,m){
      if(us <= u[i]){ 
        ru = 1; us += d[idx];
      }else{
        ru = 0; us -= d[idx];
      }
      if(vs <= v[i]){
        rd = 1; vs += d[idx];
      }else{
        rd = 0; vs -= d[idx];
      }
      if(ru){
        if(rd){
          cout << "R";
        }else{
          cout << "U";
        }
      }else{
        if(rd){
          cout << "L";
        }else{
          cout << "D";
        }
      }
    }
    cout << "\n";
  }

}