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
// ===================================
//   author: M_sea
//   website: http://m-sea-blog.com/
// ===================================
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int read() {
    int X=0,w=1; char c=getchar();
    while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
    while (c>='0'&&c<='9') X=X*10+c-'0',c=getchar();
    return X*w;
}

const int N=1000+10;
const char ch[4]={'L','R','D','U'};
const char dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};

int n,x[N],y[N];
vector<int> c;

signed main() {
    n=read();
    for (int i=1;i<=n;++i) x[i]=read(),y[i]=read();
    for (int i=2;i<=n;++i)
        if ((abs(x[i]+y[i])&1)!=(abs(x[1]+y[1])&1)) {
            puts("-1"); return 0;
        }
    for (int i=30;~i;--i) c.push_back(1<<i);
    if (!(abs(x[1]+y[1])&1)) c.push_back(1);
    printf("%lu\n",c.size());
    for (int i:c) printf("%d ",i); puts("");
    for (int i=1;i<=n;++i) {
        ll nx=0,ny=0;
        for (int j:c) {
            vector<pair<ll,int> > o;
            for (int d=0;d<4;++d) {
                ll tx=nx+dx[d]*j,ty=ny+dy[d]*j;
                o.push_back(make_pair(abs(x[i]-tx)+abs(y[i]-ty),d));
            }
            sort(o.begin(),o.end());
            int d=o.begin()->second;
            putchar(ch[d]),nx+=dx[d]*j,ny+=dy[d]*j;
        }
        puts("");
    }
    return 0;
}