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 <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include<algorithm>
#include<sstream>
#include<iomanip>


using namespace std;

typedef long long ll;

const ll MOD_CONST = 1000000007;
const ll BIG_NUM = 1000000000000000000;

const string LDUR = "LDUR";
int main() {
	int n;
	cin >> n;
	int mod2 = -1;
	bool canConsist = true;
	
	vector<int> u(n);
	vector<int> v(n);
	for (int i = 0; i < n; i++) {
		int x, y;
		cin >> x >> y;

		if (abs(x + y) % 2 != mod2 && mod2 >= 0) {
			canConsist = false;
			break;
		}
		else if (mod2 < 0) {
			mod2 = abs(x + y) % 2;
		}

		u[i] = x + y;
		v[i] = x - y;
		
	}


	if (!canConsist) {
		cout << -1 << endl;
	}
	
	else {

		if (mod2 == 0) {
			cout << 32 << endl;
		}
		else {
			cout << 31 << endl;
		}
		
		for (int i = 30; i >=0;i--) {
			cout << (1 << i) << " ";
		}
		if (mod2 ==0) {
			cout << 1;
		}
		cout << endl;
		for (int i = 0; i < n;i++) {
			ll ui = (u[i] + (1LL << 31) - 1) / 2;
			ll vi = (v[i] + (1LL << 31) - 1) / 2;
			if (mod2 == 0) {
				ui = (u[i] + (1LL << 31)) / 2;
				vi = (v[i] + (1LL << 31)) / 2;
			}
			string ans;
			for (int i = 30; i  >=0;i--) {
				//printf("%x\n", ui);
				int ub = (ui >> i) & 1;
				//cout << ub << endl;
				int vb = (vi >> i) & 1;
				//cout << ub << endl;
				ans += LDUR[ub * 2 + vb];
			}
			if (mod2 == 0) {
				ans += LDUR[0];
			}

			cout << ans << endl;
		}
	}

}
	
