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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>

#define int long long
#define MOD 1000000007

#define rep(i, n) for (int i = 0; i < (n); i++)
#define REP(i, a, n) for (int i = (a); i <= (n); i++)
#define all(a) (a).begin(), (a).end()
#define mp(a, b) make_pair((a), (b))

using namespace std;

int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, -1, 0, 1 };

signed main() {
	int N;
	cin >> N;

	vector<int> x(N), y(N);
	rep(i, N) cin >> x[i] >> y[i];

	rep(i, N) {
		if ((x[i] + y[i] + 11234567890) % 2 != (x[0] + y[0] + 11234567890) % 2) {
			cout << -1 << endl;
			return 0;
		}
	}

	vector<int> d;
	rep(i, 39) {
		d.push_back(1LL << i);
	}
	reverse(all(d));

	bool even = false;
	if ((x[0] + y[0]) % 2 == 0) {
		d.push_back(1);
		even = true;
	}

	cout << d.size() << endl;
	for (int num : d) cout << num << " ";
	cout << endl;

	rep(i, N) {
		int tx = even;
		int ty = 0;
		rep(j, 39) {
			int dx = x[i] - tx;
			int dy = y[i] - ty;
			if (abs(dx) > abs(dy)) {
				if (dx > 0) {
					cout << 'R';
					tx += d[j];
				} else {
					cout << 'L';
					tx -= d[j];
				}
			} else {
				if (dy > 0) {
					cout << 'U';
					ty += d[j];
				} else {
					cout << 'D';
					ty -= d[j];
				}
			}
		}
		if (even) cout << 'R';
		cout << endl;
	}
}
