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 i64 = long long;

template<typename T> T abs(T x) { return x < 0 ? -x : x; }

const int N = 10000 + 7;

int n, m;
i64 x[N], y[N];
i64 c[N];

int main() {
//	freopen("code.in", "r", stdin);
//	freopen("code.out", "w", stdout);

	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld%lld", x + i, y + i);
		if ((x[1] + y[1] - x[i] - y[i]) & 1) return printf("-1"), 0;
	}
	m = 36;
	for (int i = 36; i; --i)
		c[i] = 1ll << 36 - i;
	if ((x[1] + y[1]) % 2 == 0) c[++m] = 1;
	printf("%d\n", m);
	for (int i = 1; i <= m; ++i)
		printf("%lld ", c[i]);
	printf("\n");
	for(int i = 1; i <= n; ++i) {
		i64 u = 0, v = 0;
		for (int j = 1; j <= m; ++j) {
			i64 dx = abs(u - x[i]), dy = abs(v - y[i]);
			if (dx > dy)
				x[i] > u ? (printf("R"), u += c[j]) : (printf("L"), u -= c[j]);
			else
				y[i] > v ? (printf("U"), v += c[j]) : (printf("D"), v -= c[j]);
		}
		printf("\n");
	}
	return 0;
}
