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

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.

Examples

Input

3
-1 0
0 3
2 -1


Output

2
1 2
RL
UU
DR


Input

5
0 0
1 0
2 0
3 0
4 0


Output

-1


Input

2
1 1
1 1


Output

2
1 1
RU
UR


Input

3
-7 -3
7 3
-3 -7


Output

5
3 1 4 1 5
LRDUL
RDULR
DULRD
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(x) x.begin(),x.end()
#define dbg(x) cout<<#x<<":"<<x<<endl
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,P> PP;

int main(){
  ll n;
  cin>>n;
  vector<ll> x(n),y(n);
  int f[2]={};
  for(int i=0;i<n;i++){
    cin>>x[i]>>y[i];
    f[(x[i]+y[i])%2]=1;
  }
  if(f[0]&&f[1]){
    cout<<"-1"<<endl;
    return 0;
  }

  cout<<34<<endl;
  ll d[35];
  d[0]=1;
  for(int i=0;i<34;i++){
    if(i)cout<<" ";
    if(f[0]&&i==33)d[i]--;
    cout<<d[i];
    d[i+1]=d[i]*2;
  }
  cout<<endl;

  for(int i=0;i<n;i++){
    ll u=x[i]+y[i];
    ll v=x[i]-y[i];
    //    cout<<u<<" "<<v<<endl;
    int us[34],vs[34];
    if(u>0){
	 fill_n(us,34,0);
	 us[33]=1;
    }else{
	 fill_n(us,34,1);
	 us[33]=0; 
    }

    if(v>0){
	 fill_n(vs,34,0);
	 vs[33]=1;
    }else{
	 fill_n(vs,34,1);
	 vs[33]=0; 
    }
    u=abs(u);
    v=abs(v);
    u/=2;
    v/=2;
    
    int cnt=0;
    while(u||v){
	 if(u%2)us[cnt]=(us[cnt]+1)%2;
	 if(v%2)vs[cnt]=(vs[cnt]+1)%2;
	 u/=2;
	 v/=2;
	 cnt++;
    }
    
    for(int i=0;i<34;i++){
	 //	 cout<<us[i]<<" "<<vs[i];
	 if(us[i]==0&&vs[i]==0)cout<<"L";
	 if(us[i]==1&&vs[i]==1)cout<<"R";
	 if(us[i]==0&&vs[i]==1)cout<<"D";
	 if(us[i]==1&&vs[i]==0)cout<<"U";
	 //	 cout<<endl;
    }
    cout<<endl;
    
  }
  return 0;
}
