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 <string>
#define ll long long
using namespace std;

ll leng[32];
int r;

void fill_leng(){
    leng[30]=leng[31]=1;
    for(int i=29; i>=0; --i){
        leng[i]=leng[i+1]*2;
    }
}

string solve(ll X, ll Y){
    string res="";
    if(r==0) --Y;
    ll xnow=0, ynow=0;
    for(int i=0; i<31; ++i){
        ll dx=X-xnow, dy=Y-ynow;
        if(dx>dy){
            if(-dx<dy){
                res += "R";
                xnow += leng[i];
            }
            else{
                res += "D";
                ynow -= leng[i];
            }
        }
        else{
            if(-dx<dy){
                res += "U";
                ynow += leng[i];
            }
            else{
                res += "L";
                xnow -= leng[i];
            }
        }
    }
    if(r==0) res += "U";
    return res;
}

int main() {
	int N;
	cin >> N;
	vector<ll> x(N), y(N);
	for(int i=0; i<N; ++i){
	    cin >> x[i] >> y[i];
	}
	r=abs(x[0]+y[0])%2;
	for(int i=1; i<N; ++i){
	    if(abs(x[i]+y[i])%2!=r){
	        cout << -1 << endl;
	        return 0;
	    }
	}
	fill_leng();
	cout << 32-r << endl;
	cout << leng[0];
	for(int i=1; i<32-r; ++i) cout << ' ' << leng[i];
	cout << endl;
	for(int i=0; i<N; ++i) cout << solve(x[i], y[i]) << endl;
	return 0;
}
