output_description: 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.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      sx sy tx ty
attempt: 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()
  
problem_description: 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.
samples: [{"input": "0 0 1 2", "output": "UURDDLLUUURRDRDDDLLU\n    \n\nOne possible shortest path is:\n\n  * Going from (sx,sy) to (tx,ty) for the first time: (0,0) \u2192 (0,1) \u2192 (0,2) \u2192 (1,2)\n  * Going from (tx,ty) to (sx,sy) for the first time: (1,2) \u2192 (1,1) \u2192 (1,0) \u2192 (0,0)\n  * Going from (sx,sy) to (tx,ty) for the second time: (0,0) \u2192 (-1,0) \u2192 (-1,1) \u2192 (-1,2) \u2192 (-1,3) \u2192 (0,3) \u2192 (1,3) \u2192 (1,2)\n  * Going from (tx,ty) to (sx,sy) for the second time: (1,2) \u2192 (2,2) \u2192 (2,1) \u2192 (2,0) \u2192 (2,-1) \u2192 (1,-1) \u2192 (0,-1) \u2192 (0,0)\n\n* * *"}, {"input": "-2 -2 1 1", "output": "UURRURRDDDLLDLLULUUURRURRDDDLLDL"}]