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 N 1005
#define ll long long
#define debug(a) cout<<#a<<": ";for(auto i:a)cout<<i<<" ";cout<<endl;
#define trace(a) cout<<#a<<": "<<a<< endl;

int x[N],y[N];
string ans[N];
ll val[41];
string solve(ll x, ll y) {
	string str = "";
	for(int i = 31;i >= 0;i--) {
		if(abs(x) < abs(y)) {
			if(y < 0) {
				y += val[i];
				str.push_back('D');
			}
			else {
				y -= val[i];
				str.push_back('U');
			}
		}
		else {
			if(x < 0) {
				x += val[i];
				str.push_back('L');
			}
			else {
				x -= val[i];
				str.push_back('R');
			}
		}
	}
	reverse(str.begin(),str.end());
	return str;
}
int main() {
	ios_base::sync_with_stdio(false);cout.tie(0);cin.tie(0);
	int i,j,k,l,m,n,t;
	cin >> n;
	bool mark[2] = {false};
	for(i = 0;i < n;i++) {
		cin >> x[i] >> y[i];
		mark[abs(x[i]+y[i])&1] = true;
	}
	if(mark[0] && mark[1]) {
		cout << "-1" << endl;
		return 0;
	}
	vector<ll> v;
	if(mark[1]) {
		for(i = 0;i <= 31;i++) {
			val[i] = (1LL<<i);
			v.push_back(val[i]);
		}
		for(i = 0;i < n;i++) {
			ans[i] = solve(x[i],y[i]);
		}
	}
	else {
		for(i = 0;i <= 31;i++) {
			val[i] = (1LL<<i);
			v.push_back(val[i]);
		}
		for(i = 0;i < n;i++) {
			ans[i] = solve(x[i]-1,y[i]) + "R";
		}
		v.push_back(1);
	}
	cout << v.size() << endl;
	for(ll it : v) {
		cout << it << " ";
	}
	cout << endl;
	for(i = 0;i < n;i++) {
		cout << ans[i] << endl;
	}
}