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 <string>
#include <algorithm>
#define N 1234
#define int long long
template<typename T> inline void read(T &x) {
	x = 0; char c = getchar(); bool flag = false;
	while (!isdigit(c)) { if (c == '-') flag = true; c = getchar(); }
	while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }
	if (flag)	x = -x;
}
using namespace std;
inline int F(int x) {
	return x < 0 ? -x : x;
}
int n;
int X[N], Y[N];
int d[43], dtot;
inline void work(int x, int y) {
	int nx = 0, ny = 0;
	for (int i = 1; i <= dtot; ++i) {
		int t = d[i];
		if (F(x - nx) > F(y - ny)) {
			if (x < nx)	putchar('L'), nx -= t;
			else	putchar('R'), nx += t;
		} else {
			if (y < ny)	putchar('D'), ny -= t;
			else putchar('U'), ny += t;
		}
	}
	putchar('\n');
}
signed main() {
	read(n);
	int type = -1;
	for (int i = 1; i <= n; ++i) {
		read(X[i]), read(Y[i]);
		int t = (X[i] + Y[i]) % 2;
		if (t < 0)	t += 2;
		if (type == -1)	type = t;
		else if (type != t) { puts("-1"); exit(0); }
	}
	if (type)	puts("31");
	else	puts("32");
	for (int i = 30; ~i; --i)	printf("%d ", 1 << i);
	if (type == 0)	puts("1");
	else	puts("");
	int nw = 30;
	if (type == 0) {
		for (int i = 0; i <= 30; ++i)	d[++dtot] = 1 << nw, --nw;
		d[++dtot] = 1;
	} else {
		for (int i = 0; i <= 30; ++i)	d[++dtot] = 1 << nw, --nw;
	}
	for (int i = 1; i <= n; ++i)	work(X[i], Y[i]);
	return 0;
}
