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++14 (GCC 5.4.1) (judge verdict: Accepted)

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#pragma warning(disable:4786)
#pragma warning(disable:4996)
#include <ctime>
#include<list>
#include <numeric>
#include<bitset>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<functional>
#include<string>
#include<cstring>
#include<cstdlib>
#include<queue>
#include<utility>
#include<fstream>
#include<sstream>
#include<cmath>
#include<stack>
#include<assert.h>
#include<unordered_map>
#include<unordered_set>
#include <array>
using namespace std;

#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
  enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; };
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug {
#ifdef LOCAL
	~debug() { cerr << endl; }
	template<class c> typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
		cerr << boolalpha << i; return *this;
	}

	template<class c, int=0> typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
		return *this << range(begin(i), end(i));
	}

sim, class b dor(pair < b, c > d) {
	ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d) {
	*this << "[";
	for (auto it = d.b; it != d.e; ++it)
		*this << ", " + 2 * (it == d.b) << *it;
	ris << "]";
}
#else
	sim dor(const c&) { ris; }
#endif
};
#define watch(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define MEM(a, b) memset(a, (b), sizeof(a))
#define CLR(a) memset(a, 0, sizeof(a))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define ABS(X) ( (X) > 0 ? (X) : ( -(X) ) )
#define S(X) ( (X) * (X) )
#define SZ(V) (int )V.size()
#define FORN(i, n) for(int i = 0; i < n; i++)
#define FORAB(i, a, b) for(int i = a; i <= b; i++)
#define ALL(V) V.begin(), V.end()
#define IN(A, B, C)  ((B) <= (A) && (A) <= (C))
#define AIN(A, B, C) assert(IN(A, B, C))

//typedef int LL;
typedef long long int LL;
//typedef __int128 LLL;
typedef long long LLL;

typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef pair<double, double> PDD;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<PLL > VPL;
typedef vector<PII > VP;
typedef vector<double> VD;
typedef vector<vector<int>> VVI;
typedef long double ld;

const LL MOD = 998244353;
int sx, sy, tx, ty;

void move(int sx, int sy, int tx, int ty) {
	if (sx == tx) {
		char ch = (sy < ty ? 'U' : 'D');
		for (int i = MIN(sy, ty) + 1; i <= MAX(sy, ty); i++) printf("%c", ch);
		return;
	}
	if (sy == ty) {
		char ch = (sx < tx ? 'R' : 'L');
		for (int i = MIN(sx, tx) + 1; i <= MAX(sx, tx); i++) printf("%c", ch);
		return;
	}
	assert(0);
}

void solve(int ks) {
	scanf("%d %d %d %d", &sx, &sy, &tx, &ty);
	move(sx, sy, sx - 1, sy);
	move(sx - 1, sy, sx - 1, ty + 1);
	move(sx - 1, ty + 1, tx, ty + 1);
	move(tx, ty + 1, tx, ty);
	move(tx, ty, sx, ty);
	move(sx, ty, sx, sy);
	move(sx, sy, tx, sy);
	move(tx, sy, tx, ty);
	move(tx, ty, tx + 1, ty);
	move(tx + 1, ty, tx + 1, sy - 1);
	move(tx + 1, sy - 1, sx, sy - 1);
	move(sx, sy - 1, sx, sy);
}

int main()
{
#ifdef LOCAL
	double start_time = clock();
	freopen("C:\\Home\\ContestCodes\\sample.in", "r", stdin);
	//	freopen("out.out", "w", stdout);
#endif


	if (0) {
		int T;
		scanf("%d", &T);
		//AIN(T, 1, 25);
		for (int ks = 1; ks <= T; ks++) {
			solve(ks);
			fprintf(stderr, "%d done\n", ks);
		}
	}
	else {
		solve(0);
	}

#ifdef LOCAL
	double end_time = clock();
	fprintf(stderr, "Time = %lf\n", (end_time - start_time) / CLOCKS_PER_SEC);
#endif

	return 0;
}