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>
#define N 1005

using namespace std;

int n, len;
long long x[N], y[N], d[N];

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%lld %lld", &x[i], &y[i]);
	int flag = x[1] + y[1] & 1;
	for(int i = 1; i <= n; ++i)
		if((x[i] + y[i] & 1) != flag) {
			puts("-1");
			return 0;
		}

	if(!flag) d[++len] = 1;
	d[++len] = 1;
	for(int i = 1; i <= 32; ++i) {
		++len;
		d[len] = d[len - 1] << 1;
	}
	reverse(d + 1, d + len + 1);
	
	printf("%d\n", len);
	for(int i = 1; i <= len; ++i)
		printf("%lld%c", d[i], " \n"[i == len]);
	
	for(int i = 1; i <= n; ++i) {
		long long x = 0, y = 0;
		for(int j = 1; j <= len; ++j) {
			long long dL = abs((x - d[j]) - ::x[i]) + abs(y - ::y[i]);
			long long dR = abs((x + d[j]) - ::x[i]) + abs(y - ::y[i]);
			long long dU = abs(x - ::x[i]) + abs((y + d[j]) - ::y[i]);
			long long dD = abs(x - ::x[i]) + abs((y - d[j]) - ::y[i]);
			long long mic = min(min(dL, dR), min(dU, dD));
			if(dL == mic) {
				putchar('L');
				x -= d[j];
			} else if(dR == mic) {
				putchar('R');
				x += d[j];
			} else if(dU == mic) {
				putchar('U');
				y += d[j];
			} else if(dD == mic) {
				putchar('D');
				y -= d[j];
			}
		} putchar('\n');
	}
	return 0;
}