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
#pragma G++ optimize (2)
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f, mod=1000000007;
int x[1010],y[1010],sum[1010], a[1010];

void read(int &x)
{
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}

int main()
{
    int n; read(n); int maxn=0;
    for (int i=1; i<=n; i++) 
        read(x[i]), read(y[i]), 
        sum[i]=abs(x[i]+y[i]), maxn=max(sum[i],maxn);
    for (int i=1; i<=n-1; i++) 
        if ((sum[i]&1)!=(sum[i+1]&1)) {printf("-1\n"); return 0;}
    int tmp=1, cnt=0; if (!(sum[1]&1)) a[++cnt]=1, maxn--;
    while (tmp-1<=maxn)
    {
        a[++cnt]=tmp; tmp*=2;
    }
    printf("%d\n",cnt);
    for (int i=cnt; i>=1; i--) printf("%d ",a[i]); printf("\n");
    for (int i=1; i<=n; i++)
    {
        int nx=0, ny=0;
        for (int j=cnt; j>=1; j--)
        {
            int tx=x[i]-nx, ty=y[i]-ny;
            if (abs(tx)>abs(ty))
                if (tx>0) printf("R"), nx+=a[j];
                    else printf("L"), nx-=a[j];
            else
                if (ty>0) printf("U"), ny+=a[j];
                    else printf("D"), ny-=a[j];
        }
        printf("\n");
    }
    
    return 0;
}