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<sstream>
#include<vector>
#include<iterator>
#include<algorithm>
#include<climits>
using namespace std;

const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const string dir = "URDL";

string solve(long long x, long long y, const vector<long long>& arm)
{
	string ret;
	int id;
	long long d, dd, xx, yy;
	for(auto it=arm.rbegin(); it!=arm.rend(); it++){
		d = LLONG_MAX;
		for(int i=0; i<4; i++){
			xx = x + dx[i] * *it;
			yy = y + dy[i] * *it;
			dd = abs(xx) + abs(yy);

			if(dd < d){
				id = i;
				d = dd;
			}
		}

		ret += dir[(id+2)%4];
		x += dx[id] * *it;
		y += dy[id] * *it;
	}

	reverse(ret.begin(), ret.end());

	return ret;
}

int main(void)
{
	int n;
	cin >> n;

	vector<long long> x(n), y(n);
	for(int i=0; i<n; i++)
		cin >> x[i] >> y[i];

	bool b = (abs(x[0]) + abs(y[0])) % 2;
	for(int i=0; i<n; i++){
		if((abs(x[i]) + abs(y[i]))%2 != b){
			cout << -1 << endl;
			return 0;
		}
	}

	long long md = 0;
	for(int i=0; i<n; i++)
		md = max(md, abs(x[i])+abs(y[i]));

	vector<long long> arm;
	if(b == 0)
		arm.push_back(1);
	for(long long i=1, sum=(b==0 ? 1 : 0); sum<md; i*=2){
		arm.push_back(i);
		sum += i;
	}

	cout << arm.size() << endl;
	for(int i=0; i<arm.size(); i++)
		cout << arm[i] << (i+1==arm.size() ? '\n' : ' ');

	for(int i=0; i<n; i++)
		cout << solve(x[i], y[i], arm) << endl;

	return 0;
}
