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>
using namespace std;

long long x[1009], y[1009], rx[1009], ry[1009];

void rp(long long d, long long r)
{
	if (d == 0 && r == -1) cout << "R";
	else if (d == 0 && r == 1) cout << "L";
	else if (d == 1 && r == -1) cout << "U";
	else cout << "D";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int n; cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> x[i] >> y[i];
		rx[i] = ry[i] = 1;
		if (x[i] < 0) {
			x[i] *= -1; rx[i] = -1;
		}
		if (y[i] < 0) {
			y[i] *= -1; ry[i] = -1;
		}
	}
	long long t = (x[0] + y[0]) % 2;
	for (int i = 0; i < n; i++) {
		if ((x[i] + y[i]) % 2 != t) {
			cout << -1 << '\n';
			return 0;
		}
	}
	cout << 35 + t << '\n';
	long long mr = (1LL << 32LL);
	cout << mr << " ";
	if (t == 1) cout << 1 << " ";
	for (int i = 0; i <= 32; i++)
		cout << (1LL << (i * 1LL)) << " ";
	cout << (1LL << 33LL) - 1LL << '\n';
	for (int i = 0; i < n; i++) {
		y[i] += mr; rp(1, ry[i]);
		if (t == 1) {
			if (x[i] % 2 == 1) { x[i]--; rp(0, rx[i] * -1); }
			else { y[i]--; rp(1, ry[i] * -1); }
		}
		for (int j = 0; j <= 32; j++) {
			long long ti = (1LL << (j * 1LL));
			bool f = false;
			if ((x[i] & (ti * 2)) != (y[i] & (ti * 2)))
				f = true;
			if ((x[i] & ti) != 0) {
				if (f) {
					x[i] += ti; rp(0, rx[i]);
				}
				else {
					x[i] -= ti; rp(0, rx[i] * -1);
				}
			}
			else {	
				if (f) {
					y[i] -= ti; rp(1, ry[i] * -1);
				}
				else {
					y[i] += ti; rp(1, ry[i]);
				}
			}
		}
		y[i] -= (1LL << 33LL) - 1LL; rp(1, ry[i] * -1);
		cout << '\n';
	}
	return 0;
}