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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
//#define int ll
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector< vvi > vvvi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef pair<ld, ld> pld;
typedef complex<double> base;

#define mmin(a, b) a = min(a, (b))
#define mmax(a, b) a = max(a, (b))
#define mp(a,b) make_pair(a,b)
#define modsum(a, b) ((a) + (b) >= M ? (a) + (b) - M : (a) + (b))
#define modsubtract(a, b) ((a) - (b) < 0 ? (a) - (b) + M : (a) - (b))
#define modmultiply(a, b) (((a) * (b)) % M)
#define pb(a) push_back(a)
#define ALL(a) a.begin(),a.end()
#define sqr(x) (x) * (x)
#define fori(i, n) for(int i = 0; i < (int)(n); ++i)
#define cint const int &
#define SZ(a) ((int)(a.size()))
#define f(cnt) (((cnt) + 1) * (cnt) / 2)
#define watch(x) cout << (#x) << " = " << x << endl;

const double PI = 2 * acos(0.0);

template<typename T>
struct edge {
    int from, to;
    T w;
    edge(int from, int to, T w): from(from), to(to), w(w) { }
    edge() { }
};


template<class T>
void show(T a) {
    for (auto x : a)
        cout << x << " ";
    cout << endl;
}

const int N = 22;

vector<string> t = {"L", "R", "D", "U"};

inline void build(int sh, int x, string &s) {
    if (x > 0) sh++;
    x = abs(x);
    for (int i = 0; i < x; ++i)
        s += t[sh];
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin >> n;

    vector<pii> points(n);
    int mx = 0;
    fori (i, n) {
        cin >> points[i].first >> points[i].second;
        mmax(mx, abs(points[i].first) + abs(points[i].second));
    }
    fori (i, n) {
        if ((mx - abs(points[i].first) - abs(points[i].second)) & 1) {
            cout << "-1";
            return 0;
        }
    }
    vi ans(mx, 1);
    vector<string> out;
    fori (i, n) {
        string cur = "";
        build(0, points[i].first, cur);
        build(2, points[i].second, cur);
        while (cur.size() < mx) {
            cur += "LR";
        }
        assert(cur.size() == mx);
        out.push_back(cur);
    }
    cout << mx << "\n";
    for (auto x : ans)
        cout << x << " ";
    cout << endl;
    for (auto x : out)
        cout << x << "\n";
}