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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main()
{
	int N;
	cin >> N;
	vector<pair<ll, ll>> point(N);

	int md;
	for(int i(0);i<N;++i){
		cin >> point[i].first >> point[i].second;
		if(i == 0) md = (abs(point[i].first) + abs(point[i].second)) % 2;
		if(md != (abs(point[i].first) + abs(point[i].second)) % 2){
			cout << -1 << endl;
			return 0;
		}
	}
	vector<ll> d;
	for(int i(32);i>=0;--i) d.push_back(pow(2,i));
	if(md == 0) d.push_back(1);
	
	cout << d.size() << endl;
	for(int i(0);i<(d.size()-1);++i){
		cout << d[i] << ' ';
	}
	cout << d[d.size()-1] << endl;

	for(int i(0);i<N;++i){
		ll x_sum(0), y_sum(0);
		for(int j(0);j<d.size();++j){
			ll u = point[i].first + point[i].second;
			ll v = point[i].first - point[i].second;
			if(x_sum <= u && y_sum <= v){
				x_sum += d[j]; y_sum += d[j]; cout << 'R'; 
			}else if(x_sum <= u && y_sum > v){
				x_sum += d[j]; y_sum -= d[j]; cout << 'U'; 
			}else if(x_sum > u && y_sum <= v){
				x_sum -= d[j]; y_sum += d[j]; cout << 'D'; 
			}else{
				x_sum -= d[j]; y_sum -= d[j]; cout << 'L'; 
			}
		}
		cout << endl;
	}
	return 0;
}