Problem:
Score : 300 points

Problem Statement

Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up.

Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1.

Here, both the x- and y-coordinates before and after each movement must be integers.

He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy).

Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty).

Under this condition, find a shortest path for him.

Constraints

-1000 ≤ sx < tx ≤ 1000

-1000 ≤ sy < ty ≤ 1000

sx,sy,tx and ty are integers.

Input

The input is given from Standard Input in the following format:

sx sy tx ty

Output

Print a string S that represents a shortest path for Dolphin.

The i-th character in S should correspond to his i-th movement.

The directions of the movements should be indicated by the following characters:

U: Up

D: Down

L: Left

R: Right

If there exist multiple shortest paths under the condition, print any of them.

Sample Input 1

0 0 1 2

Sample Output 1

UURDDLLUUURRDRDDDLLU

One possible shortest path is:

Going from (sx,sy) to (tx,ty) for the first time: (0,0) → (0,1) → (0,2) → (1,2)

Going from (tx,ty) to (sx,sy) for the first time: (1,2) → (1,1) → (1,0) → (0,0)

Going from (sx,sy) to (tx,ty) for the second time: (0,0) → (-1,0) → (-1,1) → (-1,2) → (-1,3) → (0,3) → (1,3) → (1,2)

Going from (tx,ty) to (sx,sy) for the second time: (1,2) → (2,2) → (2,1) → (2,0) → (2,-1) → (1,-1) → (0,-1) → (0,0)

Sample Input 2

-2 -2 1 1

Sample Output 2

UURRURRDDDLLDLLULUUURRURRDDDLLDL

Solution in C (GCC 5.4.1) (judge verdict: Accepted)


#define _USE_MATH_DEFINES
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>  
#include <stdbool.h>   
#define inf (INT_MAX-1)
#define INF 9223372036854775807;
#define PI 3.14159265358979323846;
#define EPS 1e-10
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
/*
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r)
*/
#define chsort(s,n) qsort(s,n,sizeof(char),cmp)
#define chsort_r(s,n) qsort(s,n,sizeof(char),char_cmp_r)
#define TYPE long long
#define ll long long
#define MEMSET(a) memset(a,0,sizeof(a))
#define MEMSET_U(a) memset(a,-1,sizeof(a))
#define bool    _Bool
const int mod = (int)1e09 + 7;

//#define DEBUG1
//#define DEBUG2
//#define DEBUGF
#define DUMMY

int in(void) { int i; scanf("%d", &i); return i; }
long long llin(void) { long long i; scanf("%lld", &i); return i; }
double din(void) { double i; scanf("%lf", &i); return i; }
void chin(char s[]) { scanf("%s", s); }

void print(int a) { printf("%d\n", a); }
void llprint(long long a) { printf("%lld\n", a); }
void dprint(double a) { printf("%.10f\n", a); }
void print2(int a, int b) { printf("%d %d\n", a, b); }
int Max(int a, int b) { if (a > b) { return a; }return b; }
int Min(int a, int b) { if (a < b) { return a; }return b; }
long long llmax(long long a, long long b) { return a > b ? a : b; }
long long llmin(long long a, long long b) { return a < b ? a : b; }
double dmax(double a, double b) { return a > b ? a : b; }
double dmin(double a, double b) { return a < b ? a : b; }
//long long llmax(long long a, long long b) { return a > b ? a : b; }
//long long llmin(long long a, long long b) { return a < b ? a : b; }
//double dmax(double a, double b) { return a > b ? a : b; }
int cmp(const void *a, const void *b) { return *(TYPE *)a - *(TYPE *)b; }
int cmp_r(const void *a, const void *b) { return *(TYPE *)b - *(TYPE *)a; }
int char_cmp(const void *a, const void *b) { return strcmp((char *)a, (char *)b); }
int char_cmp_r(const void *a, const void *b) { return strcmp((char *)b, (char *)a); }
void swap(int *a, int *b) { int t = *a; *a = *b; *b = t; }



int main() {
	int sx, sy, tx, ty;
	scanf("%d%d%d%d", &sx, &sy, &tx, &ty);

	int dx = (tx - sx);
	int dy = (ty - sy);

	//path1
	for (int i = 0; i < dx; i++) { printf("R"); }
	for (int i = 0; i < dy; i++) { printf("U"); }
	//path2
	for (int i = 0; i < dx; i++) { printf("L"); }
	for (int i = 0; i < dy; i++) { printf("D"); }
	//path3
	printf("D");
	for (int i = 0; i < dx+1; i++) { printf("R"); }
	for (int i = 0; i < dy+1; i++) { printf("U"); }
	printf("L");
	//path4
	printf("U");
	for (int i = 0; i < dx + 1; i++) { printf("L"); }
	for (int i = 0; i < dy + 1; i++) { printf("D"); }
	printf("R");
	printf("\n");

#ifdef DEBUGF
	getch();
#endif
	return 0;
}