Problem: 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.

Input

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.

Output

If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 ≤ l < r ≤ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.

Examples

Input

8
1 6 5 4 3 2 7 8


Output

2 6


Input

4
2 3 4 1


Output

0 0


Input

4
1 2 3 4


Output

0 0

Solution (in JAVA):

Solution:
'use strict'

let lll
let N = +readline()
lll = readline().split(' ').map(v => parseInt(v))
let ss = []
for (let i = 1; i < lll.length; i++) {
  ss.push(Math.sign(lll[i] - lll[i - 1]))
}

let ifm = ss.indexOf(-1)
let ilm = ss.lastIndexOf(-1)

if (~ifm && ~ilm) {
  let bad
  for (let i = ifm; i < ilm; i++) {
    if (ss[i] == 1) {
      bad = true
      break
    }
  }
  if (bad) {
    print('0 0')
  } else {
    let ns = lll.slice(0, ifm).concat(lll.slice(ifm, ilm + 2).reverse(), lll.slice(ilm + 2))
    for (let i = 1; i < ns.length; i++) {
      if (ns[i] < ns[i - 1]) {bad = true; break}
    }
    print(bad ? '0 0' : (ifm + 1) + ' ' + (ilm + 2))
  }
} else {
  print('0 0')
}