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: Time Limit Exceeded)

#include <functional>
#include <list>
#include <string>
#include <vector>
#include <climits>
#include <cstring>
#include <map>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <set>
#include <deque>
#include <stack>
#include <cassert>
using namespace std;
#define FOR(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)
#define ALL(x) (x).begin(),(x).end()
#define CASET2 int ___T, case_n = 1; scanf("%d ", &___T); while ((___T > 0 ? printf("Case #%d:\n", case_n++) : 0), ___T-- > 0)
#define CASET1 int ___T, case_n = 1; scanf("%d ", &___T); while ((___T > 0 ? printf("Case #%d: ", case_n++) : 0), ___T-- > 0)
#define CASET int ___T; scanf("%d ", &___T); while (___T-- > 0)
#define SZ(X) ((int)(X).size())
#define LEN(X) strlen(X)
#define REP(i,n) for(int i=0;i<int(n);i++)
#define REP1(i,a,b) for(int i=(a);i<=int(b);i++)
#define PER1(i,a,b) for(int i=(a);i>=(b);i--)
#define REPL(i,x) for(int i=0;x[i];i++)
#define PER(i,n) for(int i=(n)-1;i>=0;i--)
#define RI1(x) scanf("%d",&x)
#define RI2(x,y) RI1(x), RI1(y)
#define RI3(x,y...) RI1(x), RI2(y)
#define RI4(x,y...) RI1(x), RI3(y)
#define RI5(x,y...) RI1(x), RI4(y)
#define RI6(x,y...) RI1(x), RI5(y)
#define GET_MACRO(_1, _2, _3, _4, _5, _6, NAME, ...) NAME
#define RI(argv...) GET_MACRO(argv, RI6, RI5, RI4, RI3, RI2, RI1)(argv)
#define DRI(argv...) int argv;RI(argv)
#define WRI(argv...) while (RI(argv) != EOF)
#define DWRI(x...) int x; WRI(x)
#define RS(x) scanf("%s",x)
#define MP make_pair
#define PB push_back
#define MS0(x) memset(x,0,sizeof(x))
#define MS1(x) memset(x,-1,sizeof(x))
#define X first
#define Y second
#define V(x) vector<x >

typedef long double LD;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef long long LL;
const int INF = 1000000000;

void print(int i) { printf("%d", i); }
void print(LL i);
template<class T> void PI(T i) { print(i); puts(""); }
template<class T> void PIS(T i) { print(i); printf(" "); }

template<class T>
void PV(T const &v, int N) {
    REP(i, N) {
        print(v[i]);
        printf("%c", i == N-1 ? '\n' : ' ');
    }
}

template<class T> void PV(T const &v) { PV(v, SZ(v)); }

template<class T, class S> bool has_bit(T mask, S i) { return (mask >> i) & 1; }
long long shift(int i) { return 1ll << i; }

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
int popcount(int x) { return __builtin_popcount(x); } 
int popcount(long long x) { return __builtin_popcountll(x); } 

#include <unordered_set>
#include <unordered_map>
#define EB emplace_back
#define RL(x) scanf("%lld", &(x))
#define PL(x) printf("%lld\n", x)
#define DRL(x) LL x; RL(x)

int main() {
    DRI(N);
    VI x(N), y(N);
    int m = 0;
    REP(i, N) {
        RI(x[i], y[i]);
        if (( x[i] ^ y[i] ^ x[0] ^ y[0] ) % 2) {
            PI(-1);
            return 0;
        }
        m = max(m, abs(x[i]) + abs(y[i]));
    }
    PI(m);
    REP(i, m) {
        i == m-1 ? PI(1) : PIS(1);
    }
    REP(i, N) {
        REP(j, abs(x[i]))
            printf(x[i] < 0 ? "L" : "R");
        REP(j, abs(y[i])) {
            printf(y[i] < 0 ? "D" : "U");
        }
        REP(j, (m - abs(x[i]) - abs(y[i])) / 2) {
            printf("UD");
        }
        puts("");
    }
    return 0;
}