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 Rint register int
using namespace std;
typedef long long LL;
template<typename T>
inline void read(T &x){
    int ch = getchar(); x = 0;
    bool f = false;
    for(;ch != '-' && (ch < '0' || ch > '9');ch = getchar());
    if(ch == '-'){f = true; ch = getchar();}
    for(;ch >= '0' && ch <= '9';ch = getchar()) x = x * 10 + ch - '0';
    if(f) x = -x;
}
const int N = 1003;
int n, x[N], y[N], t, ans[40], m;
char str[40];
int main(){
    read(n);
    for(Rint i = 1;i <= n;++ i) read(x[i]), read(y[i]);
    t = abs(x[1] + y[1]) & 1;
    for(Rint i = 2;i <= n;++ i)
        if((abs(x[i] + y[i]) & 1) != t) return puts("-1"), 0;
    printf("%d\n1", 32 - t);
    ans[0] = m = 1;
    if(!t) printf(" 1"), ans[m ++] = 1;
    for(Rint i = 1;i <= 30;++ i) printf(" %d", ans[m ++] = 1 << i); putchar('\n');
    for(Rint i = 1;i <= n;++ i){
        for(Rint j = m - 1, nowx = 0, nowy = 0;~j;-- j){
            int dx = x[i] - nowx, dy = y[i] - nowy;
            if(abs(dx) > abs(dy)){
                if(dx > 0) nowx += ans[j], str[j] = 'R';
                else nowx -= ans[j], str[j] = 'L';
            } else {
                if(dy > 0) nowy += ans[j], str[j] = 'U';
                else nowy -= ans[j], str[j] = 'D';
            }
        }
        printf("%s\n", str);
    }
}