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

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.

Examples

Input

3
-1 0
0 3
2 -1


Output

2
1 2
RL
UU
DR


Input

5
0 0
1 0
2 0
3 0
4 0


Output

-1


Input

2
1 1
1 1


Output

2
1 1
RU
UR


Input

3
-7 -3
7 3
-3 -7


Output

5
3 1 4 1 5
LRDUL
RDULR
DULRD
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>

using namespace std;

typedef pair<int, int> P;

int main() {
  int N;
  cin >> N;
  vector<int> X(N);
  vector<int> Y(N);
  for (int i = 0; i < N; i++) {
    cin >> X[i] >> Y[i];
  }
  vector<P> pos(N);

  int parity = abs(X[0] + Y[0]) % 2;
  for (int i = 0; i < N; i++) {
    int parity2 = abs(X[i] + Y[i]) % 2;
    pos[i] = make_pair(X[i]+Y[i], X[i]-Y[i]);
    if (parity2 != parity) {
      cout << -1 << endl;
      return 0;
    }
  }
  vector<int> ds;
  for (int i = 30; i >= 0; i--) {
    ds.push_back(1LL << i);
  }
  if (parity == 0) {
    ds.push_back(1);
  }
  cout << ds.size() << endl;
  for (auto d : ds) {
    cout << d << " ";
  }
  cout << endl;
  for (int i = 0; i < N; i++) {
    int xsum = 0;
    int ysum = 0;
    int xdir, ydir;
    for (auto d : ds) {
      if (xsum >= pos[i].first) {
        xdir = -1;
        xsum -= d;
      } else {
        xdir = 1;
        xsum += d;
      }
      if (ysum >= pos[i].second) {
        ydir = -1;
        ysum -= d;
      } else {
        ydir = 1;
        ysum += d;
      }
      if (xdir > 0 && ydir > 0) {
        cout << 'R';
      } else if (xdir > 0 && ydir < 0) {
        cout << 'U';
      } else if (xdir < 0 && ydir > 0) {
        cout << 'D';
      } else {
        cout << 'L';
      }
    }
    cout << endl;
  }
}
