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 <iostream>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <map>

#define reps(i,s,n) for(int (i) = (s); (i) < (n); (i)++)
#define rep(i,n) reps(i,0,n)
using namespace std;
using ll = long long;

int main(){
    int n;
    cin >> n;
    vector<int> x(n) ,y(n);
    cin >> x[0] >> y[0];
    int mod = abs(x[0]+y[0])%2;
    bool f = false;
    
    reps(i,1,n){
        cin >> x[i] >> y[i];
        if(abs(x[i]+y[i])%2 != mod){
            f = true;
            break;
        }
    }
    
    if(f) cout << -1 << endl;
    else{
        cout << 32 - mod << endl;
        rep(i,31) cout << (int)pow(2,i) << " ";
        if(mod == 0) cout << 1 ;
        cout << "" << endl;
        vector<bool> u(31),v(31);
        rep(i,n){
            if(mod == 0) x[i]--;
            rep(j,31)u[j] = v[j] = false;
            rep(j,31){
                if((
                        ((x[i]+y[i]-1)/2+(int)pow(2,30)) >> j
                )&1) u[j] = true;
                if((
                        ((x[i]-y[i]-1)/2+(int)pow(2,30)) >> j
                )&1) v[j] = true;
            }
            rep(j,31){
                if(u[j] && v[j]) cout << "R";
                if(u[j] && !v[j]) cout << "U";
                if(!u[j] && v[j]) cout << "D";
                if(!u[j] && !v[j]) cout << "L";
            }
            if(mod == 0) cout << "R";
            cout << endl;
        }
    }

    return 0;
}