Inna and Dima bought a table of size *n* × *m* in the shop. Each cell of the table contains a single letter: "D", "I", "M", "A".

Inna loves Dima, so she wants to go through his name as many times as possible as she moves through the table. For that, Inna acts as follows:
 1.  initially, Inna chooses some cell of the table where letter "D" is written; 1.  then Inna can move to some side-adjacent table cell that contains letter "I"; then from this cell she can go to one of the side-adjacent table cells that contains the written letter "M"; then she can go to a side-adjacent cell that contains letter "A". Then Inna assumes that she has gone through her sweetheart's name; 1.  Inna's next move can be going to one of the side-adjacent table cells that contains letter "D" and then walk on through name DIMA in the similar manner. Inna never skips a letter. So, from the letter "D" she always goes to the letter "I", from the letter "I" she always goes the to letter "M", from the letter "M" she always goes to the letter "A", and from the letter "A" she always goes to the letter "D". 
Depending on the choice of the initial table cell, Inna can go through name DIMA either an infinite number of times or some positive finite number of times or she can't go through his name once. Help Inna find out what maximum number of times she can go through name DIMA.
The first line of the input contains two integers *n* and *m* (1 ≤ *n*, *m* ≤ 10<sup class="upper-index">3</sup>). 

Then follow *n* lines that describe Inna and Dima's table. Each line contains *m* characters. Each character is one of the following four characters: "D", "I", "M", "A". 

Note that it is not guaranteed that the table contains at least one letter "D".
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
# from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
    sys.setrecursionlimit(int(pow(10,6)))
    sys.stdin = open("input.txt", "r")
    # sys.stdout = open("../output.txt", "w")
except:
    pass


n, m = L()
m += 1
q = {'I': 0, 'M': 1, 'A': 2, 'D': 3}
t = []
for i in range(n):
    t += map(q.get, input())
    t.append(-7)
t += [-7] * m
# print(t)
p = [[] for q in t]
c = [0] * len(t)
for a in range(n * m):
    for b in (a - m, a + m, a - 1, a + 1):
        if abs(t[b] - t[a] + 1) == 2:
            p[a].append(b)
            c[b] += 1
s = [i for i, q in enumerate(c) if not q]
while s:
    a = s.pop()
    for b in p[a]:
        t[b] = max(t[b], t[a] + 1)
        c[b] -= 1
        if c[b] == 0: s.append(b)
k = max(t) - 2 >> 2
print('Poor Inna!' if any(c) else k if k > 0 else 'Poor Dima!')


endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
