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
///created: 14/Jun/2019 16:07:17

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> ii;
#define db(x) cerr << "debug: " << #x << " = " << x << '\n';
template<class __Ty> bool minimize(__Ty& a, __Ty b) {
	if (a > b) { a = b; return true; }
	return false;
}
template<class __Ty> bool maximize(__Ty& a, __Ty b) {
	if (a < b) { a = b; return true; }
	return false;
}

const int hor[] = { 0, +1, -1, 0 };
const int ver[] = { -1, 0, 0, +1 };
const int w[] = { 'D', 'R', 'L', 'U' };
const int N = 1e3 + 5;
ll a[N], b[N], d[35];
int n, m;

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	cin >> n;
	int cnt = 0;
	for (int i = 1; i <= n; ++i) {
		cin >> a[i] >> b[i];
		cnt += int(abs(a[i] + b[i])) % 2;
	}
	if ((cnt != 0) && (cnt != n)) {
		cout << -1;
		return 0;
	}
	if (cnt == 0) {
		m = 32;
		d[1] = d[2] = 1;
		for (int i = 3; i <= m; ++i)
			d[i] = d[i - 1] * 2;
	}
	else {
		m = 31;
		d[1] = 1;
		for (int i = 2; i <= m; ++i)
			d[i] = d[i - 1] * 2;
	}
	printf("%d\n", m);
	reverse(d + 1, d + 1 + m);
	for (int i = 1; i <= m; ++i)
		printf("%lld ", d[i]);
	printf("\n");
	for (int i = 1; i <= n; ++i) {
		ll x = 0, y = 0;
		for (int j = 1; j <= m; ++j) {
			ll res = ll(1e18), t = 0;
			for (int k = 0; k < 4; ++k)
				if (minimize(res, abs(x + hor[k] * d[j] - a[i]) + abs(y + ver[k] * d[j] - b[i]))) {
					t = k;
				}
			x += 1LL * hor[t] * d[j];
			y += 1LL * ver[t] * d[j];
			printf("%c", w[t]);
		}
		printf("\n");
		fprintf(stderr, "[%lld, %lld] -> [%lld, %lld]\n", a[i], b[i], x, y);
	}
	return 0;
}
