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 ll = long long;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  
  int odd = 0, even = 0;
  vector<ll> x(n), y(n);
  for (int i = 0; i < n; i++) {
    cin >> x.at(i) >> y.at(i);
    if ((abs(x.at(i)) + abs(y.at(i))) % 2 == 1) odd++;
    else even++;
  }
  
  if (odd > 0 && even > 0) {
    cout << "-1\n";
    return 0;
  }
  
  vector<ll> arms;
  for (ll i = 35; i >= 0; i--) arms.emplace_back(1LL << i);
  if (even > 0) arms.emplace_back(1);
  
  cout << arms.size() << '\n';
  for (int i = 0; i < arms.size(); i++) {
    if (i) cout << " ";
    cout << arms.at(i);
  }
  cout << '\n';
  
  for (int i = 0; i < n; i++) {
    ll gu = x.at(i) + y.at(i), gv = x.at(i) - y.at(i);
    ll nowx = 0, nowy = 0;
    for (int j = 0; j < arms.size(); j++) {
      ll nowu = nowx + nowy, nowv = nowx - nowy;
      if (nowu <= gu && nowv <= gv) {
        cout << 'R';
        nowx += arms.at(j);
      }
      else if (nowu >= gu && nowv >= gv) {
        cout << 'L';
        nowx -= arms.at(j);
      }
      else if (nowu <= gu && nowv >= gv) {
        cout << 'U';
        nowy += arms.at(j);
      }
      else {
        cout << 'D';
        nowy -= arms.at(j);
      }
    }
    cout << '\n';
    // cout << nowx << " " << nowy << endl;
  }
}