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>
#include <algorithm>

using namespace std;
typedef long long ll;
typedef pair<long long,long long> P;
#define rep(n) for(ll i=0; i<n; i++)
#define MAX_V 100001
const ll INF = 1e+9;
const ll MOD = 1000000007;

ll pow(ll x, ll a)
{
  if(a==0) return 1;
  ll y = pow(x, a/2);
  y *= y;
  if(a%2==1) y *= x;
  return y;
}

bool bit(ll x, ll i)
{
  return (x >> i) & 1;
}

void res(const vector<ll>& D, const P& R, bool isEven)
{
  ll U = pow((ll)2, (ll)31) -1 + R.first;
  U /= 2;
  ll V = pow((ll)2, (ll)31) -1 + R.second;
  V /= 2;

  if(isEven) printf("R");

  for (ll i = 0; i < 31; i++)
  {
    if(bit(U, i))
    {
      if(bit(V,i))
      {
        printf("R");
      } else {
        printf("U");
      }
    } else {
      if(bit(V,i))
      {
        printf("D");
      } else {
        printf("L");
      }
    }
  }
  printf("\n");
}

int main()
{
  ll n;
  cin >> n;
  vector<P> pos;
  for (ll i = 0; i < n; i++)
  {
    ll x, y;
    cin >> x >> y;
    pos.emplace_back(x+y, x-y);
  }

  for (ll i = 0; i < n-1; i++)
  {
    if(abs(pos[i].first)%2 != abs(pos[i+1].first)%2)
    {
      cout << -1 << endl;
      return 0;
    }
  }

  vector<ll> D;
  D.push_back(1);
  for (ll i = 0; i < 30; i++)
  {
    D.push_back(D[i]*2);
  }
  
  bool isEven = pos[0].first % 2 == 0;
  if(isEven)
  {
    for (ll i = 0; i < n; i++)
    {
      pos[i].first--;
      pos[i].second--;
    }    
    printf("32\n");
    printf("1 ");
  } else {
    printf("31\n");
  }

  for(auto d:D) printf("%d ", d);
  printf("\n");

  for (ll i = 0; i < n; i++)
  {
    res(D, pos[i], isEven);
  }
  
  return 0;
}