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>
#define rep(i, n) for(int i = 1; i <= n; ++i)
using namespace std;
#define int long long
const int maxn = 1020;
int n, x[maxn], y[maxn], odd, even;
namespace sol{
    int a[59], k, ham;
    const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
    const char dr[] = "DULR";
    void dfs(int x, int y, int cur) {
        if(cur == 0) return;
        int chg = 1LL << (cur - 1);
        for(int dir = 0; dir < 4; ++dir) {
            int xf = x + dx[dir] * chg, yf = y + dy[dir] * chg;
            if(abs(xf) + abs(yf) < (1LL << (cur - 1)) ) {
                putchar(dr[dir]);
                dfs(xf, yf, cur - 1);
                break;
            }
        }
    }
    inline void solve() {
        rep(i, n) ham = max(ham, x[i] + y[i]) + 2;
        for(k = 0; (1LL << k) < ham; ++k) ;
        if(even) printf("%lld\n1 ", ++k + 1);
        else printf("%lld\n", ++k);
        for(int i = k; i ; --i) {
            a[i] = 1LL << (i - 1);
            printf("%lld%c", a[i], i == 1LL ? '\n' : ' ');
        }
        for(int i = 1; i <= n; ++i){
            if(even) putchar('L'), x[i]++;
            dfs(x[i], y[i], k);
            putchar('\n');
        }
    }
}
signed main(){
    scanf("%lld", &n);
    odd = even = true;
    rep(i, n) scanf("%lld%lld", &x[i], &y[i]);
    rep(i, n) {
        int sum = abs(x[i]) + abs(y[i]);
        if(sum & 1) even = false; else odd = false;
    }
    if((odd | even) == 0) {puts("-1"); return 0;}
    sol::solve();
    return 0;
}
////