Problem:
Score : 600 points

Problem Statement

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

Partial Score

In the test cases worth 300 points, -10 \leq X_i \leq 10 and -10 \leq Y_i \leq 10 hold.

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.

Sample Input 1

3
-1 0
0 3
2 -1

Sample Output 1

2
1 2
RL
UU
DR

In the given way to bring Joint m of the robot arm to each (X_j, Y_j), the positions of the joints will be as follows:

To (X_1, Y_1) = (-1, 0): First, the position of Joint 0 is (x_0, y_0) = (0, 0). As the mode of Section 1 is R, the position of Joint 1 is (x_1, y_1) = (1, 0). Then, as the mode of Section 2 is L, the position of Joint 2 is (x_2, y_2) = (-1, 0).

To (X_2, Y_2) = (0, 3): (x_0, y_0) = (0, 0), (x_1, y_1) = (0, 1), (x_2, y_2) = (0, 3).

To (X_3, Y_3) = (2, -1): (x_0, y_0) = (0, 0), (x_1, y_1) = (0, -1), (x_2, y_2) = (2, -1).

Sample Input 2

5
0 0
1 0
2 0
3 0
4 0

Sample Output 2

-1

Sample Input 3

2
1 1
1 1

Sample Output 3

2
1 1
RU
UR

There may be duplicated points among (X_j, Y_j).

Sample Input 4

3
-7 -3
7 3
-3 -7

Sample Output 4

5
3 1 4 1 5
LRDUL
RDULR
DULRD

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define int long long //data is large
#define rep(i, _st, _ed) for(register int i = (_st); i <= (_ed); i++)
int n, x[1005], y[1005], m, arm[55]; 
char str[55];
const int dx[] = {-1, 1, 0, 0},
dy[] = {0, 0, -1, 1} ;
const char op[] =  {'L', 'R', 'D', 'U'};
inline int read(){int ans = 0, f = 1; char c = getchar();while(c < '0' || c > '9') f = (c == '-') ? -1 : f, c = getchar();while('0' <= c && c <= '9') ans = ans*10 + c - '0', c = getchar();return ans;}
int dis(int x, int y) {
	return abs(x) + abs(y);
}
signed main(){
	n = read();
	rep(i, 1, n) x[i] = read(), y[i] = read();
	int parity = (x[1] + y[1]) % 2, has_sol = 1;
	rep(i, 1, n) if( (x[i] + y[i]) % 2 != parity) has_sol = 0;
	if(has_sol) {
		m = 3, arm[1] = 1, arm[0] = 1;
		rep(i, 2, m) arm[i] = 2*arm[i-1];
		
		printf("%lld\n", m + (parity == 0));
		if(!parity) printf("1 ");
		rep(i, 1, m) printf("%lld ", arm[i]);
		puts("");
		
		rep(i, 1, n) {
			memset(str, 0, sizeof str);
			int nx, ny;
			for(int j = m; j >= 0; j--) {
				if(parity && j == 0) break;
				rep(k, 0, 3) {
					nx = x[i] + dx[k]*arm[j];
					ny = y[i] + dy[k]*arm[j];
					if((nx == 0 && ny == 0) || dis(nx, ny) <= (1 << (j-1))) {
						x[i] = nx;
						y[i] = ny;
						str[j] = op[k];
						break;
					}
				}

			}
			
			if(!parity) {
				printf("%s\n", str);
			} else printf("%s\n", str+1);
		} 
	} else puts("-1"); 
	return 0;
}



/*
3
-1 0
0 3
2 -1

*/