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 <algorithm>
#include <cstdio>
#include <functional>
#include <iostream>
#include <cfloat>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <map>
#include <unordered_map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <vector>
using namespace std;

int N, X[2000], Y[2000], odds = 0;
void findPath(int x, int y) {
  for(int i = 30;i >= 0;i--) {
    if((abs(x) < abs(y)) & y > 0) {
      cout << "U";
      y = y - (1 << i);
    } else if((abs(x) < abs(y)) & y < 0) {
      cout << "D";
      y = y + (1 << i);
    } else if((abs(x) > abs(y)) & x > 0) {
      cout << "R";
      x = x - (1 << i);
    } else {
      cout << "L";
      x = x + (1 << i);
    }
  }
}

int main() {
  //freopen("input.in","r",stdin);
  //freopen("output.out","w",stdout);

  cin >> N;
  for(int i = 0;i < N;i++) {
    cin >> X[i] >> Y[i];
    if((abs(X[i]) + abs(Y[i])) % 2 != 0) odds++;
  }

  if(odds > 0 and odds < N) {
    cout << "-1" << endl;
    return 0;
  }

  if(odds == 0) {
    for(int i = 0;i < N;i++) X[i] -= 1;
  }

  if(odds == 0) cout << "32" << endl;
  else cout << "31" << endl;

  if(odds == 0) cout << 1 << " ";
  for(int i = 30;i >= 0;i--) {
    cout << (1LL << i) << " ";
  }

  cout << endl;

  for(int i = 0;i < N;i++) {
    if(odds == 0) cout << "R";
    findPath(X[i], Y[i]);
    cout << endl;
  }

  return 0;
}
