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>
using namespace std;
#define ALL(A) (A).begin(),(A).end()
#define ll long long
const ll mod = 1e9+7;
const ll INF = 2*1e18;
const int inf = 1e9+7;

ll n,dsize;
vector<ll> x,y,d;
string dir ="LUDR";

string solve (int now);
int main(void){
	cin >> n;
	x.resize(n);
	y.resize(n);
	for(int i=0;i<n;i++)cin >> x[i] >> y[i];
	for(int i=1;i<n;i++){
		if(abs(x[i-1] + y[i-1])%2 != abs(x[i]+y[i])%2){
			cout << -1 << endl;
			return 0;
		}
	}
	for(int i=0;i<31;i++)d.push_back( ( 1LL << (30-i) ) );
	if(!(abs(x[0]+y[0])%2))d.push_back(1);
	dsize = d.size();
	cout << dsize << endl;
	for(int i = 0;i < dsize;i++){
		if(i != 0 )cout << " ";
		cout << d[i];
	}
	cout << endl;
	for(int i=0;i<n;i++)cout << solve(i) << endl;
	return 0;
}

string solve(int now){
	string ans;
	ll uv[2] = {x[now]+y[now]+(1LL<<31)-1,x[now]-y[now]+(1LL<<31) -1 };
	if(abs(x[now]+y[now])%2==0)++uv[0],++uv[1];
	for(int i=0;i<dsize;i++){
		int flag = 0;
		for(int j=0;j<2;j++){
			if(uv[j] >=2*d[i]){
				flag += 1 << j;
				uv[j] -= 2*d[i];
			}
		}
		ans += dir[flag];
	}
	assert(uv[0] == 0 && uv[1] == 0 );
	return ans;
}