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 <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <map>
#include <set>
#include <complex>
#include <cmath>
using namespace std;


int main(int argc, char const *argv[]) {
    int n;cin>>n;
    vector<long long> x(n),y(n);
    for(int i=0;i<n;++i){
        cin>>x[i]>>y[i];
    }
    for(int i=1;i<n;++i){
        if((x[i]+y[i]-x[0]-y[0])%2!=0){
            cout<<-1<<endl;
            return 0;
        }
    }
    int m=33;
    vector<long long> d(m+2);
    long long ll = 1;
    for(int i=0;i<m;++i){
        d[i] = ll;
        ll<<=1;
    }
    bool par = (x[0]+y[0])%2 == 0;
    d[m] = (1ll<<(m-1)) + par;
    d[m+1] = (1ll<<m);
    vector<string> res(n);
    for(int i=0;i<n;++i){
        string s = "";
        long long u = x[i] + y[i] + (1ll<<(m-1)) + par, v = x[i] - y[i] + (1ll<<(m-1)) + par;
        for(int i=0;i<m;++i){
            u>>=1;
            v>>=1;
            bool b=u%2,c=v%2;
            if(b){
                if(c)s+='R';
                else s+='U';
            }
            else{
                if(c)s+='D';
                else s+='L';
            }
        }
        s+="LR";
        res[i] = s;
    }
    cout << d.size() << endl;
    for(long long l:d)cout<<l<<" ";
    cout<<endl;
    for(string s:res)cout<<s<<endl;
    return 0;
}