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<bits/stdc++.h>

using namespace std;

using int64 = long long;

int main() {
  int64 N, X[1000], Y[1000];
  cin >> N;
  int type = -1;
  for(int i = 0; i < N; i++) {
    int x, y;
    cin >> x >> y;

    int t = (x + y) % 2 == 0 ? 0 : 1;
    if(type == -1) type = t;
    if(type != t) {
      cout << -1 << endl;
      return 0;
    }

    X[i] = x - y;
    Y[i] = x + y;
  }

  vector< int64 > use;
  int64 sum = 0;
  for(int64 i = 31; i >= 0; i--) {
    use.emplace_back(1LL << i);
    sum += 1LL << i;
  }

  cout << use.size() + (type == 0) << endl;
  for(int i = 0; i < use.size(); i++) {
    cout << use[i] << " ";
  }

  if(type == 0) {
    cout << 1;
    for(int j = 0; j < N; j++) {
      X[j]--;
      Y[j]--;
    }
  }
  cout << endl;


  for(int i = 0; i < N; i++) {
    int64 latte = (X[i] + sum) / 2;
    int64 malta = (Y[i] + sum) / 2;
    for(int64 j = 31; j >= 0; j--) {
      int t1 = (latte >> j) & 1;
      int t2 = (malta >> j) & 1;
      if(t1 == 0 && t2 == 0) cout << "L";
      else if(t1 == 1 && t2 == 1) cout << "R";
      else if(t1 == 1 && t2 == 0) cout << "D";
      else cout << "U";
    }
    if(type == 0) cout << "R";
    cout << endl;
  }

}