Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define eb emplace_back
#define mt make_tuple
#define all(x) (x).begin(), (x).end() 
#define MOD 1000000007

typedef long long ll;
typedef pair <int, int> ii;
typedef pair <ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef long double ld;

const ll INF=LLONG_MAX;
const int mxn=1001;
pll cor[mxn];
int dx[]={-1,0,1,0};
int dy[]={0,-1,0,1};
char d[]={'R','U','L','D'};

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    //freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
	int n; cin >> n;
	int odd = 0;
	for(int i=0; i<n; i++){
		ll x,y;
		cin >> x >> y;
		cor[i].fi = x;
		cor[i].se = y;
		if((x+y)%2)odd++;
	}
	vll bt;
	if(odd==n){ // all odd
		bt.eb(1LL);
		for(int i=1; i<31; i++)bt.eb(bt.back()*2LL);
	}else if(odd==0){ // all even
		bt.eb(1LL);
		bt.eb(1LL);
		for(int i=1; i<31; i++)bt.eb(bt.back()*2LL);
	}else{
		cout << -1 << endl;
		return 0;
	}
	int m = bt.size();
	cout << m << endl;
	for(auto x:bt)cout << x << ' '; cout << endl;
	for(int i=0; i<n; i++){
		ll x = cor[i].fi;
		ll y = cor[i].se;
		string s="";
		for(int j=m-1; j>=0; j--){
			for(int k=0; k<4; k++){
				ll X = x+(ll)dx[k]*bt[j];
				ll Y = y+(ll)dy[k]*bt[j];
				if(ll(abs(X)+abs(Y))<=(ll)bt[j]){
					s+=d[k];
					x = X; y = Y;
					break;
				}
			}
		}
		reverse(all(s));
		cout << s << endl;
	}
}
	
// READ & UNDERSTAND
// ll, int overflow, array bounds, memset(0)
// special cases (n=1?), n+1 (1-index)
// do smth instead of nothing & stay organized
// WRITE STUFF DOWN