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>
#define endl '\n'
#define debug(X) cout << #X << " = " << X << endl

using namespace std;

const int MX = 30;

int pt2[MX];

int check(int x, int y, int k) {
    return abs(x) + abs(y) <= pt2[k] - 1;
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	#ifdef UTP
	freopen("input.txt", "r", stdin);
	#endif
	int n; cin >> n;
	vector<int> x(n), y(n);
	int ok = 1;
	for (int i = 0; i < n; i++) {
        cin >> x[i] >> y[i];
        ok &= ((abs(x[i]) + abs(y[i])) % 2 == (abs(x[0]) + abs(y[0])) % 2);
	}
    if (!ok) {cout << -1 << endl; return 0; }
    int even = (abs(x[0]) + abs(y[0])) % 2 == 0;
    for (int i = 0; i < MX; i++)
        pt2[i] = (1 << i);
    cout << MX + even << endl;
    if (even) cout << 1 << " ";
    for (int i = MX - 1; i >= 0; i--)
        cout << pt2[i] << " ";
    cout << endl;
    for (int j = 0; j < n; j++) {
        string ans;
        int cx = x[j], cy = y[j];
        if (even) {
            ans.push_back('R');
            cx--;
        }
        for (int i = MX - 1; i >= 0; i--) {
            if (check(cx, cy + pt2[i], i)) {
                ans.push_back('D');
                cy += pt2[i];
            } else if (check(cx + pt2[i], cy, i)) {
                ans.push_back('L');
                cx += pt2[i];
            } else if (check(cx, cy - pt2[i], i)) {
                ans.push_back('U');
                cy -= pt2[i];
            } else if (check(cx - pt2[i], cy, i)) {
                ans.push_back('R');
                cx -= pt2[i];
            } else {
                assert(false);
            }
        }
        assert(cx == 0 && cy == 0);
        cout << ans << endl;
    }
	return 0;
}