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 <iostream>
#include <sstream>
#include <stdio.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <utility>
#include <functional>
#include <cmath>

#define REP(i,x,y) for(int i=(x);i<(y);i++)
#define RREP(i,x,y) for(int i=(y)-1;i>=(x);i--)
#define all(x) (x).begin(),(x).end()
// #define int long long

using namespace std;

// conversion
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define dump(x)  cerr << #x << " = " << (x) << endl
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl
#else
#define eprintf(...) 42
#define dump(x) 42
#define debug(x) 42
#endif

typedef long long ll;

const int MAX_N = 1000;
int N;
int X[MAX_N];
int Y[MAX_N];

void solve() {
    int mod2 = (abs(X[0]) + abs(Y[0])) % 2;
    REP(i,0,N) {
        if (mod2 != (abs(X[i]) + abs(Y[i])) % 2) {
            cout << "-1" << "\n";
            return;
        }
    }

    const int M = 31;
    int m = mod2 == 1 ? M : M + 1;
    ll d[32];
    cout << m << "\n";
    REP(i,0,M) {
        d[M-i-1] = pow(2,i);
        cout << d[M-i-1];
        if (i != M-1) cout << " ";
    }
    if (mod2 == 0) {
        d[31] = 1;
        cout << " 1\n";
    } else {
        cout << "\n";
    }

    REP(i,0,N) {
        ll x = X[i], y = Y[i];
        REP(j,0,m) {
            ll L, R, U, D;
            L = abs(x - d[j]) + abs(y);
            R = abs(x + d[j]) + abs(y);
            U = abs(x) + abs(y + d[j]);
            D = abs(x) + abs(y - d[j]);
            if (min(min(min(L,R),U),D) == L) {
                x = x - d[j];
                cout << "L";
            } else if (min(min(min(L,R),U),D) == R) {
                x = x + d[j];
                cout << "R";
            } else if (min(min(min(L,R),U),D) == U) {
                y = y + d[j];
                cout << "U";
            } else {
                y = y - d[j];
                cout << "D";
            }
        }
        cout << "\n";
    }

    return;
}

signed main() {
    // to shorten execution time for iostream
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> N;
    REP(i,0,N) {
        cin >> X[i];
        cin >> Y[i];
    }
    solve();
    return 0;
}