#!/usr/bin/python3
from math import inf
from collections import deque

n, m = map(int, input().split())
field = []
for y in range(n):
    field.append(list(input().strip()))
    if "D" in field[-1]:
        start = (field[-1].index("D"), y)
    if "U" in field[-1]:
        end = (field[-1].index("U"), y)
    if "P" in field[-1]:
        xc, yc = (field[-1].index("P"), y)

# Zjistíme si kdy bude políčko zabrané karnevalem
carnival = [[inf]*m for i in range(n)]
direction = (-1, 0)
# Po 4*n*m (princip holubníku) krocích karneval navštíví stejné políčko
# ve stejném směru, takže bude chodit dál dokola 
for time in range(4*n*m):
    if not (0 <= xc < m and 0 <= yc < n):
        break  # Karneval vylezl z plochy
    carnival[yc][xc] = min(time, carnival[yc][xc])
    
    for i in range(4):
        ny, nx = yc + direction[0], xc + direction[1]
        if field[ny][nx] in "_P":
            break
        direction = (direction[1], -direction[0])
    
    if i == 4:
        break  # Karneval se zasekl

    xc, yc = nx, ny

# Uděláme bfs
queue = deque([(start, None, 0)])
visited = [[None]*m for i in range(n)]
while len(queue):
    (x, y), last, time = queue.popleft()
    if field[y][x] == "X" or visited[y][x] or carnival[y][x] <= time:
        continue
    visited[y][x] = last
    direction = (1, 0)
    for i in range(4):
        ny, nx = y + direction[0], x + direction[1]
        if 0 <= nx < m and 0 <= ny < n:
            queue.append(((nx, ny), (x, y), time+1))
        direction = (direction[1], -direction[0])

# Zrekonstruujeme cestu
directions = {
    (0, +1): 'D',
    (0, -1): 'N',
    (+1, 0): 'P',
    (-1, 0): 'L',
}
result = ""
last = end
while last != start:
    this = visited[last[1]][last[0]]
    result += directions[(last[0] - this[0], last[1] - this[1])]
    last = this

print(result[::-1])