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 <stdio.h>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <iomanip>
#include <string>
#include <queue>
#include <numeric>
#include <functional>
#include <array>
using namespace std;

using pll = pair<long long, long long>;

int main() {
	int N; cin >> N;
	vector<pll> pos(N);
	int par;
	for (int i = 0; i < N; ++i) {
		cin >> pos[i].first >> pos[i].second;
		if (i == 0)par = (abs(pos[0].first) + abs(pos[0].second)) % 2;
		else if(par != ((abs(pos[i].first) + abs(pos[i].second)) % 2)){
			cout << -1; return 0;
		}
	}
		
	vector<long long>deg;
	for (int i = 30; 0 <= i; --i)deg.push_back(1LL << i);
	if (!par)deg.push_back(1);
	cout << deg.size() << endl;
	for (int i = 0; i < deg.size(); ++i)cout << deg[i] << " ";
	cout << endl;

	constexpr array<long long, 4> dx = { -1, +1, 0, 0 };
	constexpr array<long long, 4> dy = { 0, 0, -1, +1 };

	for (int i = 0; i < N;++i) {
		long currentX = 0;
		long currentY = 0;
		string dirLetter = "LRDU";
		for (auto&& d : deg) {
			int dl = 0;
			long long maxDistance = numeric_limits<long long>::max();
			for (int j = 0; j < 4;++j) {
				long nextX = currentX + d * dx[j];
				long nextY = currentY + d * dy[j];
				long distance = abs(nextX-pos[i].first) + abs(nextY-pos[i].second);
				if (distance < maxDistance) {
					maxDistance = distance;
					dl = j;
				}
			}
			cout << dirLetter[dl];
			currentX += d * dx[dl];
			currentY += d * dy[dl];
		}
		cout << endl;
	}
}
