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 input = `2
2 1`.split('\n');
function print(output) {
	console.log(output);
}
function readline() {
	return input.shift();
}*/

//code

var n = Number(readline());
var num = readline().split(' ').map(Number);
num.unshift(0);
var j = 1,
    lIndex = void 0,
    rIndex = void 0,
    sum = void 0;
while (j < num.length && num[j] === j) {
	j++;
}
//first break
lIndex = j;
if (lIndex === num.length || num[lIndex] <= lIndex || num[lIndex] > n) {
	print("0 0");
} else {
	sum = lIndex + num[lIndex];
	rIndex = num[lIndex];
	while (j <= rIndex && num[j] + j === sum) {
		j++;
	}
	//second break
	if (j !== rIndex + 1) {
		print("0 0");
	} else {
		while (j < num.length && num[j] === j) {
			j++;
		}
		//last break
		if (j === num.length) {
			print(lIndex + " " + rIndex);
		} else {
			print("0 0");
		}
	}
}