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 <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std ;

#define ll long long

const int N = 2010 ;

int n ;
int cnt ;
int _eve ;
ll nx, ny ;
int bc[N] ;
struct rg{
    int l ;
    int r ;
    int t ;
}base[N] ; //D  R  U  L
int dx[4] = {0, 1, 0, -1} ;
int dy[4] = {-1, 0, 1, 0} ;
char s[4] = {'U', 'L', 'D', 'R'} ;

int main(){
    cin >> n ;
    bc[0] = 1 ;
    for (int i = 1 ; i <= n ; ++ i){
        cin >> base[i].l >> base[i].r ;
        base[i].t = abs(base[i].l + base[i].r) & 1 ;
    }
    for (int i = 2 ; i <= n ; ++ i)
        if (base[i].t != base[1].t) return puts("-1"), 0 ;
    for (int i = 1 ; i <= 30 ; ++ i) bc[++ cnt] = 1 << i ;
    cout << cnt + 1 + (bool)(!(base[1].t & 1)) << '\n' ;
    if (!(base[1].t & 1)) cout << 1 << " " ;
    for (int i = cnt ; ~i ; -- i) printf("%d%c", bc[i], " \n"[!i]) ;
    for (int i = 1 ; i <= n ; ++ i){
        nx = base[i].l ;
        ny = base[i].r ;
        if (!(base[i].t & 1))
            ny -= 1, putchar('U') ;
		//cout << nx << " " << ny << endl ;
		for (int j = cnt ; ~j ; -- j){
            for (int k = 0 ; k < 4 ; ++ k){
                ll kx = nx + (ll)dx[k] * bc[j] ;
                ll ky = ny + (ll)dy[k] * bc[j] ;
                if (abs(kx) + abs(ky) < bc[j]){
                    nx = kx ; ny = ky ;
					//cout << nx << " " << ny << endl ;
                    printf("%c", s[k]) ; break ;
                }
			}
		}
        puts("") ;
    }
    return 0 ;
}
