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 Python (3.4.3) (judge verdict: Accepted)

import sys
import math
import collections
import itertools
import array
import inspect

# Set max recursion limit
sys.setrecursionlimit(1000000)


# Debug output
def chkprint(*args):
    names = {
        id(v): k
        for k, v in inspect.currentframe().f_back.f_locals.items()
    }
    print(', '.join(
        names.get(id(arg), '???') + ' = ' + repr(arg) for arg in args))


# Binary converter
def to_bin(x):
    return bin(x)[2:]


def li_input():
    return [int(_) for _ in input().split()]


def gcd(n, m):
    if n % m == 0:
        return m
    else:
        return gcd(m, n % m)


def gcd_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = gcd(v, L[i])

    return v


def lcm(n, m):
    return (n * m) // gcd(n, m)


def lcm_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = lcm(v, L[i])

    return v


# Width First Search (+ Distance)
def wfs_d(D, N, K):
    """
    D: 隣接行列(距離付き)
    N: ノード数
    K: 始点ノード
    """

    dfk = [-1] * (N + 1)
    dfk[K] = 0

    cps = [(K, 0)]
    r = [False] * (N + 1)
    r[K] = True
    while len(cps) != 0:
        n_cps = []
        for cp, cd in cps:
            for i, dfcp in enumerate(D[cp]):
                if dfcp != -1 and not r[i]:
                    dfk[i] = cd + dfcp
                    n_cps.append((i, cd + dfcp))
                    r[i] = True

        cps = n_cps[:]

    return dfk


# Depth First Search (+Distance)
def dfs_d(v, pre, dist):
    """
    v:  現在のノード
    pre: １つ前のノード
    dist: 現在の距離

    以下は別途用意する
    D: 隣接リスト(行列ではない)
    D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
    """

    global D
    global D_dfs_d

    D_dfs_d[v] = dist

    for next_v, d in D[v]:
        if next_v != pre:
            dfs_d(next_v, v, dist + d)

    return


def sigma(N):
    ans = 0
    for i in range(1, N + 1):
        ans += i
    return ans


def comb(n, r):
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n

    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]

    for p in range(2, r + 1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p - 1, r, p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot

    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])

    return result

def bisearch(L, target):
    low = 0
    high = len(L) - 1
    
    while low <= high:
        mid = (low + high) // 2
        guess = L[mid]
        if guess == target:
            return True
        elif guess < target:
            low = mid + 1
        elif guess > target:
            high = mid - 1
    if guess != target:
        return False

# --------------------------------------------

dp = None


def main():
    sx, sy, tx, ty = li_input()

    ans = ""
    ans += "U" * (ty - sy)
    ans += "R" * (tx - sx)
    ans += "D" * (ty - sy)
    ans += "L" * (tx - sx)
    ans += "L"
    ans += "U" * ((ty - sy) + 1)
    ans += "R" * ((tx - sx) + 1)
    ans += "D"
    ans += "R"
    ans += "D" * ((ty - sy) + 1)
    ans += "L" * ((tx - sx) + 1)
    ans += "U"

    print(ans)


main()