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 REP(i,s,n) for(int i=s;i<n;++i)
#define rep(i,n) REP(i,0,n)
#define ALL(x) x.begin(),x.end()
#define EPS (1e-8)
#define equals(a,b) (fabs((a)-(b))<EPS)

using namespace std;

typedef long long ll;
bool LTE(double a,double b) { return equals(a,b) || a < b; }
bool LT(double a,double b) { return !equals(a,b) && a < b; }

#define MAX 1010
int N;
ll X[MAX], Y[MAX];

#define PLUS true
#define MINUS false

void compute() {
  int type = ( llabs(X[0]) + llabs(Y[0]) ) & 1;
  REP(i,1,N) if( type != ( ( llabs(X[i]) + llabs(Y[i]) ) & 1 ) ) {
    puts("-1");
    return;
  }
  vector<ll> d;
  rep(i,39) d.push_back(1LL<<i);
  reverse(ALL(d));
  if( type == 0 ) d.push_back(1);
  cout << d.size() << endl;
  rep(i,(int)d.size()) {
    if( i ) printf(" ");
    cout << d[i];
  } puts("");
  rep(i,N) {
    ll u = X[i] + Y[i], v = X[i] - Y[i];
    ll x = 0, y = 0;
    rep(j,(int)d.size()) {
      bool sig1,sig2;
      if( u >= x ) sig1 = PLUS, x += d[j];
      else         sig1 = MINUS,x -= d[j];
      if( v >= y ) sig2 = PLUS, y += d[j];
      else         sig2 = MINUS,y -= d[j];

      if( sig1 == MINUS      && sig2 == MINUS ) cout << "L";
      else if( sig1 == PLUS  && sig2 == PLUS  ) cout << "R";
      else if( sig1 == MINUS && sig2 == PLUS  ) cout << "D";
      else if( sig1 == PLUS  && sig2 == MINUS ) cout << "U";
    }
    puts("");
  }

}

int main() {
  cin >> N;
  rep(i,N) cin >> X[i] >> Y[i];
  compute();
  return 0;
}
