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 pii = pair<int, int>;

template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return 1; } return 0; }

const string dirs = "RUDL";
const ll INF = 1LL << 40;
int N;
ll X[1010], Y[1010];
ll x[1010], y[1010];
int m = 32;
ll d[33];

ll len(ll sx, ll sy, ll tx, ll ty) {
  return abs(sx-tx) + abs(sy-ty);
}

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> X[i] >> Y[i];
    x[i] = X[i] - Y[i];
    y[i] = X[i] + Y[i];
  }

  for (int i = 0; i < N; i++) {
    if ((x[i] - x[0]) % 2 == 1) {
      cout << -1 << endl;
      return 0;
    }
  }
  if (x[0] % 2 == 0) m++;
  d[31] = d[m-1] = 1;
  for (int i = 30; i >= 0; i--) {
    d[i] = d[i+1] << 1;
  }

  cout << m << endl;
  for (int i = 0; i < m; i++) {
    cout << d[i] << " \n"[i+1==m];
  }

  for (int t = 0; t < N; t++) {
    ll sx = 0, sy = 0;
    for (int i = 0; i < m; i++) {
      int idx = -1;
      ll tmp = INF;
      for (int j = 0; j < 4; j++) {
        ll tx = sx + ((j >> 0 & 1) ? -d[i] : d[i]);
        ll ty = sy + ((j >> 1 & 1) ? -d[i] : d[i]);
        if (chmin(tmp, len(tx, ty, x[t], y[t]))) {
          idx = j;
        }
      }
      sx += ((idx >> 0 & 1) ? -d[i] : d[i]);
      sy += ((idx >> 1 & 1) ? -d[i] : d[i]);
      cout << dirs[idx];
    }
    cout << endl;
  }

  return 0;
}