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;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;

vector<ll> b;

string solve(ll x, ll y) {
  ll u = x + y;
  ll v = x - y;
  string ret;

  string dir = "LDUR";
  for (int i = 30; i >= 0; i--) {
    int j = 0;
    if (u < 0) {
      j += 0;
      u += b[i];
    } else {
      j += 2;
      u -= b[i];
    }
    if (v < 0) {
      j += 0;
      v += b[i];
    } else {
      j += 1;
      v -= b[i];
    }
    ret.push_back(dir[j]);
  }
  reverse(ret.begin(), ret.end());

  return ret;
}

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  b.resize(31);
  b[0] = 1;
  for (int i = 0; i < 30; i++) b[i + 1] = b[i] * 2;
  int n;
  cin >> n;
  vector<int> x(n), y(n);
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
    mp[abs(x[i] + y[i]) % 2]++;
  }
  if (mp[0] != n && mp[0] != 0) {
    cout << -1 << endl;
    return 0;
  }
  if (mp[0] > 0) {
    cout << 32 << endl;
    ll m = 1;
    cout << 1;
    for (int i = 0; i < 31; i++) {
      cout << " " << m;
      m <<= 1;
    }
    cout << endl;
    for (int i = 0; i < n; i++) {
      cout << "R";
      cout << solve(x[i] - 1, y[i]) << endl;
    }
  } else {
    cout << 31 << endl;
    ll m = 2;
    cout << 1;
    for (int i = 0; i < 30; i++) {
      cout << " " << m;
      m <<= 1;
    }
    cout << endl;
    for (int i = 0; i < n; i++) {
      cout << solve(x[i], y[i]) << endl;
    }
  }

  return 0;
}
