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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
std::mt19937 rnd(time(NULL));
#define rand rnd
#define pr std::pair<int,int>
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
template<class T>void cxk(T&a,T b){a=a>b?a:b;}
template<class T>void cnk(T&a,T b){a=a<b?a:b;}
#ifdef mod
int pow(int x,int y){
	int ret=1;
	while(y){
		if(y&1)ret=1ll*ret*x%mod;
		x=1ll*x*x%mod;y>>=1;
	}
	return ret;
}
template<class Ta,class Tb>void inc(Ta&a,Tb b){a=a+b>=mod?a+b-mod:a+b;}
template<class Ta,class Tb>void dec(Ta&a,Tb b){a=a>=b?a-b:a+mod-b;}
#endif
ll len[41],x[1001],y[1001];
int main(){
#ifdef LOCAL
	freopen("in.in","r",stdin);
	//freopen("out.out","w",stdout);
#endif
	int n=gi(),s=0;
	for(int i=1;i<=n;++i)x[i]=gi(),y[i]=gi(),s|=1<<(x[i]+y[i]&1);
	if(s==3)puts("-1"),exit(0);
	int m=0;len[++m]=1;if(s==1)len[++m]=1;
	for(int i=1;i<=31;++i)len[++m]=1ll<<i;
	std::reverse(len+1,len+m+1);
	printf("%d\n",m);for(int i=1;i<=m;++i)printf("%lld ",len[i]);puts("");
	for(int i=1;i<=n;++i){
		ll X=0,Y=0;
		for(int j=1;j<=m;++j){
			ll dL=llabs(X-len[j]-x[i])+llabs(Y-y[i]);
			ll dR=llabs(X+len[j]-x[i])+llabs(Y-y[i]);
			ll dU=llabs(X-x[i])+llabs(Y+len[j]-y[i]);
			ll dD=llabs(X-x[i])+llabs(Y-len[j]-y[i]);
			ll mn=std::min({dL,dR,dU,dD});
			if(mn==dL)X-=len[j],putchar('L');
			else if(mn==dR)X+=len[j],putchar('R');
			else if(mn==dU)Y+=len[j],putchar('U');
			else Y-=len[j],putchar('D');
		}
		assert(X==x[i]&&Y==y[i]);
		puts("");
	}
	return 0;
}

