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 <string>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <numeric>
#include <cmath>

using namespace std;

typedef long long int ll;

#define all(x) x.begin(),x.end()

const ll mod = 1e9+7;
const ll INF = 1e9;
const ll MAXN = 1e9;

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

	int m = 31;
	vector<ll> d(m+1);
	for(int i = 0; i <= m; i++){
		d[i] = pow(2,m-i);
	}
	if(u[0] %2 == 0){
		d.push_back(1);
	}

	vector<ll> arm;
	vector<string> ans(n);
	for(int i = 0; i < n; i++){
		ll uu = 0,vv = 0;
		//cout << u[i] << " " << v[i] << endl;
		for(int j = 0; j < d.size(); j++){
			if(uu<u[i]){
				uu += d[j];
				if(vv<v[i]){
					vv += d[j];
					ans[i] += 'R';
				}else{
					vv -= d[j];
					ans[i] += 'U';
				}
			}else{
				 uu -= d[j];
				 if(vv<v[i]){
					vv += d[j];
					ans[i] += 'D';
				}else{
					vv -= d[j];
					ans[i] += 'L';
				}
			}
			//cout << uu << " " << vv << endl;
		}

		if(uu != u[i] || vv != v[i]){
			cout << -1 << endl;
			return 0;
		}
		// printf("\n");
	}


	cout << d.size() << endl;
	for(int i = 0; i < d.size(); i++){
		cout << d[i] << (i!=d.size()-1?" ":"\n");
	}
	for(int i = 0; i < n; i++){
		cout << ans[i] << endl;
	}

	return 0;
}