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;
template<class T> struct rge { T b, e; };
template<class T> rge<T> range(T i, T j) { return rge<T>{i, j}; }
struct debug {
#ifdef LOCAL
	~debug(){cerr<<endl;}
	template<class T> debug& operator<<(T x){cerr<<boolalpha<<x; return *this;}
	template<class B, class C> debug& operator<<(pair<B, C> x){return *this<<"("<<x.first<<", "<<x.second<<")";}
	template<class T> debug& operator<<(rge<T> x){*this<<"[";for(auto it=x.b;it!=x.e;it++){*this<<", "+2*(it==x.b)<<*it;}return *this<<"]";}
	template<class T> debug& operator<<(vector<T> x){ return *this<<range(x.begin(),x.end());}
#else
	template<class T> debug& operator <<(const T&){return *this;}
#endif
};
#define nav(...) << "[ " << #__VA_ARGS__ ": " << (__VA_ARGS__) << " ] "
using ll = long long;

const int BITS = 31;

struct P {
    int x, y;
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    /*  all points must have same parity of manhattan dist 
     * 
     * 
     *  to get to point (x, y):
     *          
     * 
     *          m = L + R + U + D (m <= 40)
     *
     */
    int n;
    cin >> n;
    vector<P> a(n);
    for(int i = 0; i < n; i++) {
        cin >> a[i].x >> a[i].y;
    }
    
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(((a[i].x+a[i].y)%2+2)%2 != ((a[j].x+a[j].y)%2+2)%2) {
                cout << "-1\n";
                exit(0);
            }
        }
    }
    
    bool even = (a[0].x + a[0].y) % 2 == 0;
    
    vector<int> d(BITS);
    for(int i = 0; i < BITS; i++) {
        d[i] = 1<<i;
    }
    if(even) {
        d.push_back(1);
    }
    int m = d.size();
    cout << m << '\n';
    for(int i = 0; i < m; i++) {
        cout << d[i] << " \n"[i==m-1];
    }
    
    for(int i = 0; i < n; i++) {
        P p = a[i];
        if(even) {
            p.x--;
        }
        string ans;
        int x = p.x, y = p.y;
        for(int b = BITS-1; b >= 0; b--) {
            int take = 1<<b;
            int need = (1<<b)-1;
            debug() nav(x) nav(y) nav(need);
            if(abs(x-take) + abs(y) <= need) {
                ans.push_back('R');
                x -= take;
            } else if(abs(x+take) + abs(y) <= need) {
                ans.push_back('L');
                x += take;
            } else if(abs(x) + abs(y-take) <= need) {
                ans.push_back('U');
                y -= take;
            } else if(abs(x) + abs(y+take) <= need) {
                ans.push_back('D');
                y += take;
            }
            else {
                assert(false); // :( so sad
            }
        }
        reverse(ans.begin(), ans.end());
        if(even) {
            ans.push_back('R');
        }
        cout << ans << '\n';
    }
    
}