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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include "bits/stdc++.h"

#include <unordered_set>
#include <unordered_map>

//#include <iostream>//POJ
//#include <vector>
//#include <string>
//#include <iomanip>
//#include <math.h>
//#include <algorithm>
//#include <cstring>
using namespace std;
#define setc cin.tie(0);ios::sync_with_stdio(0)
#define dd(n) cout<<fixed<<setprecision(n)
#define inp(x) (cin>>x,x)
#define ind(x) (cin>>x,--x)
//#define repi(i,a,b) for(int i=(a), i##_len=(b); i<i##_len; ++i)
#define repi(i,a,b) for(int i=(a);i<b;++i)
#define rep(i,n) repi(i,0,n)
#define repi_(i,a,b) for(int i=(a), i##_len=(b); i<=i##_len; ++i)
#define rep_(i,n) repi_(i,0,n)
#define repir(i,a,b) for(int i=(a)-1, i##_first=(b); i>=i##_first; --i)
#define repr(i,n) repir(i,n,0)
#define repir_(i,a,b) for(int i=(a), i##_first=(b); i>=i##_first; --i)
#define repr_(i,n) repir_(i,n,0)
#define prep(i,n) for(int i=0;i*i<n;++i)
#define prepi(i,a,b) for(int i=a;i*i<b;++i)
#define all(x) (x).begin(),(x).end()
#define vsort(v) sort((v).begin(), (v).end())
#define vcsort(v,c) sort((v).begin(),(v).end(),c)
#define vrev(v) reverse((v).begin(), (v).end())
#define fnd(v,x) ((v).find(x)!=string::npos)
#define mod(a,b) (a<0?a%b+abs(b):a%b)
#define cl(a,b) (a+b-1/b)

//template<typename T> T vpop(vector<T> *v) {
//	T d = v->back();
//	v->pop_back();
//	return d;
//}
//char vpop(string *v) {
//	char d = v->back();
//	v->pop_back();
//	return d;
//}
#define siz(v) ((int)(v).size())
#define ers(v, n) (v).erase((v).begin() + n)
#define cnt(v, n) count(all(v), n)
#define vmin(v) *min_element(v.begin(), v.end())
#define vmax(v) *max_element(v.begin(), v.end())
#define contain(q) !q.empty()
#define cont(q) !q.empty()
//#define qpop(q, a, b) a=q.back().first;b=q.back().second;q.pop()
//#define pqpop(p, a, b) a=q.top().first;b=q.top().second;q.pop()
#define el "\n"
#define sp " "
#define pi 3.14159265358979
#define co(x) cout<<x<<el
#define coc(c, a, b) co(((c)?a:b))
#define cim(x) {cin>>x;--x;}
#define cim2(a,b) {cin>>a>>b;--a;--b;}
#define cosp(x) cout<<(x)<<' '
#define YES(c) coc(c,"YES", "NO")
#define Yes(c) coc(c,"Yes", "No")
#define yes(c) coc(c,"yes", "no")
#define POSSIBLE(c) coc(c, "POSSIBLE", "IMPOSSIBLE")
#define Possible(c) coc(c, "Possible", "Impossible")
#define possible(c) coc(c, "possible", "impossible")
#define inf INT_MAX
#define wildcard(T) numeric_limits<T>::min()
#define noway() {co(-1);gc();return 0;}

#define pb push_back
#define pq priority_queue
#define np next_permutation

#ifdef tqktmp_2
#define gc() getchar(); getchar()
#else
#define gc() 1
#endif
const unsigned int bf0 = (1 << 0);
const unsigned int bf1 = (1 << 1);
const unsigned int bf2 = (1 << 2);
const unsigned int bf3 = (1 << 3);
const unsigned int bf4 = (1 << 4);
const unsigned int bf5 = (1 << 5);
const unsigned int bf6 = (1 << 6);
const unsigned int bf7 = (1 << 7);
const int _10j9p7 = 1000000007;

//#define lint long long
typedef long long lint;
typedef vector<int> IV; typedef vector<string> SV;
typedef vector<lint> LIV;
typedef vector<vector<int>> IVV;
typedef pair<int,int> P; typedef pair<lint,lint> LP;
typedef vector<P> PV; typedef vector<LP> LPV;

const int
dx8[8] ={ 0,1,1,1,0,-1,-1,-1 },
dy8[8] ={ 1,1,0,-1,-1,-1,0,1 },
dx9[9] ={ 0,1,1,1,0,0,-1,-1,-1 },
dy9[9] ={ 1,1,0,-1,0,-1,-1,0,1 },
dx5[5] ={ 0,1,0,0,-1 },
dy5[5] ={ 1,0,0,-1,0 },
dx4[4] ={ 0,1,0,-1 },
dy4[4] ={ 1,0,-1,0 };
//template<class... A> void cim_(A... args) {
//	for (A *i : initializer_list<A*>{ args... }) {
//		cin>>*i;--*i;
//	}
//	return;
//}

template<class T>inline bool maxi(T &a,const T &b){ if(a < b){ a = b; return 1; } return 0; }
template<class T>inline bool mini(T &a,const T &b){ if(b < a){ a = b; return 1; } return 0; }

#define MAX_N 1000
const int md = 1e9+7;

int main(){
	setc;

	lint n;cin>>n;
	LIV x(n),y(n);cin>>x[0]>>y[0];
	bool mo=(x[0]+y[0])%2;
	
	repi(i,1,n){
		cin>>x[i]>>y[i];
		if((abs(x[i])+abs(y[i]))%2!=mo){
			co(-1);gc();return 0;
		}
	}co(40-mo);
	rep(i,39-mo)cout<<1<<sp;co(1);
	lint ctr;
	rep(i,n){
		ctr=0;
		rep(j,abs(x[i]))cout<<(++ctr,x[i]>=0?'R':'L');
		rep(j,abs(y[i]))cout<<(++ctr,y[i]>=0?'U':'D');
		rep(i,(40-mo-ctr)/2)cout<<"UD";
		cout<<el;
	}
	gc();
}