Vasya collects coins: he has exactly one coin for every year from 1 to *n*. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from *l* to *r* inclusively and put them in the reverse order. That is, he took a certain segment [*l*, *r*] and reversed it. At that the segment's endpoints did not coincide. For example, if *n* = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... *n* using exactly one segment reversal. If it is possible, find the segment itself.
The first line contains an integer *n* (1 ≤ *n* ≤ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated *n* integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to *n*, and every number is used exactly 1 time.
n = int(input())
currentPermutation = input().split()
for i in range(0, n):
    currentPermutation[i] = int(currentPermutation[i])

copy = currentPermutation
if copy == copy.sort(reverse=True):
    print(f"1 {n}")
else:
    l = 0
    r = 0
    for i in range(0, n - 1):
        if currentPermutation[i] + 1 == currentPermutation[i + 1]:
            continue
        else:
            l = i + 1 + 1
            break

    reversedPermutation = currentPermutation
    reversedPermutation.reverse()

    for i in range(0, n - 1):
        if currentPermutation[i] - 1 == currentPermutation[i + 1]:
            continue
        else:
            r = n - i - 2
            break

    # print(l, " ", r)

    arr = [currentPermutation[i] for i in range(l, r + 1)]
    copy = arr
    arr.reverse()
    con = 1
    for i in range(0, l-r+1):
        if arr[i] + 1 == arr[i + 1]:
            continue
        else:
            con = 0
            break

    if con == 1:
        print(f"{l} {r}")
    else:
        print("0 0")
