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 <string>
#include <vector>
#include <algorithm>
#include <cstdlib>
 
using namespace std;
using lli = long long int;
 
int main() {
    int N;
    cin >> N;
    vector<lli> x(N), y(N);
    lli mx = 0;
    int odd = 0, even = 0;
    for (int i = 0; i < N; i++) {
	cin >> x[i] >> y[i];
	mx = max(mx, abs(x[i]) + abs(y[i]));
	if (abs(x[i] - y[i]) % 2) {
	    odd++;
	} else {
	    even++;
	}
    }
    if (odd * even != 0) {
	cout << -1 << endl;
	return 0;
    }
 
    vector<lli> edges;
    lli sum = 0;
    if (even) {
	edges.push_back(1);
	sum += 1;
    }
    
    for (int i = 0; sum < mx; i++) {
	edges.push_back(1LL << i);
	sum += (1LL << i);
    }
 
    cout << edges.size() << endl;
    for (int i = 0; i < edges.size(); i++) {
	cout << edges[i] << " \n"[i + 1 == edges.size()];
    }
 
 
    int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
    char ss[5] = "LDRU";
    for (int i = 0; i < N; i++) {
	vector<char> ans;
	for (int j = edges.size() - 1; j >= 0; j--) {
	    lli s = 1LL << 60;
	    int k_index = 0;
	    for (int k = 0; k < 4; k++) {
		if (abs(x[i] + dx[k] * edges[j]) + abs(y[i] + dy[k] * edges[j]) <= s) {
		    k_index = k;
		    s = abs(x[i] + dx[k] * edges[j]) + abs(y[i] + dy[k] * edges[j]);
		}
	    }
	    x[i] += dx[k_index] * edges[j];
	    y[i] += dy[k_index] * edges[j];
	    //cout << "!" << x[i] << " " << y[i] << endl;
	    ans.push_back(ss[k_index]);
	}
	for (int t = ans.size() - 1; t >= 0; t--) {
	    cout << ans[t];
	}
	cout << endl;
    }
    
    return 0;
}
