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>
#define FOR(i,k,n) for(int i=(k);i<(int)(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(x) begin(x),end(x)

using namespace std;
using ll = int64_t;

int main() {
  int n;
  cin>>n;
  vector<pair<ll, ll>> vp;
  REP(i,n) {
    ll x,y;
    cin>>x>>y;
    vp.emplace_back(x, y);
  }
  vector<string> vs(n);
  vector<ll> vl;
  string dir = "RULD";
  REP(i,40) {
    ll mx = 0;
    REP(j,n) {
      ll x, y;
      tie(x, y) = vp[j];
      mx = max(mx, abs(x) + abs(y));
    }
    if (mx == 0) {
      cout << i << endl;
      REP(j, vl.size()) {
        if (j) cout << ' ';
        cout << vl[j];
      }
      cout << endl;
      for (const auto& s : vs) {
        cout << s << endl;
      }
      return 0;
    }
    ll l = (mx + 1) / 2;
    vl.push_back(l);
    ll dx[] = {l, 0, -l, 0};
    ll dy[] = {0, l, 0, -l};
    REP(j,n) {
      ll x, y;
      tie(x, y) = vp[j];
      int idx = 0;
      ll mn = mx;
      REP(k,4) {
        ll ax = dx[k];
        ll ay = dy[k];
        ll d = abs(x-ax)+abs(y-ay);
        if (d < mn) {
          mn = d;
          idx = k;
        }
      }
      vs[j].push_back(dir[idx]);
      vp[j] = make_pair(x - dx[idx], y - dy[idx]);
    }
  }
  cout << -1 << endl;
  return 0;
}
