Polycarp is choosing three problems for creating a programming test. Totally he has $n$ problems in his list. The complexity of the $i$-th problem equals $r_i$. All problems are numerated from $1$ to $n$.
Help Polycarp to choose such three problems $a$, $b$ and $c$, so that the complexity of the first problem strictly less than the complexity of second problem and the complexity of the second problem is strictly less than the complexity of the third problem. So, for chosen problems $a$, $b$ and $c$ it should be true that $r_a < r_b < r_c$.
If Polycarp can choose three problems in different ways, you can print any of them.
Input
The first line of the input contains one integer $n$ ($3 \le n \le 3000$) — the number of problems in Polycarp’s list.
The second line of the input contains $n$ integers $r_1, r_2, \dots, r_n$ ($1 \le r_i \le 10^9$), where $r_i$ is the complexity of the $i$-th problem.
Output
If Polycarp has no ways to choose three problems, you should print three numbers -1. Ih there is a way to choose them, you should print three different integers $a, b, c$ ($1 \le a, b, c \le n$), where $a$ is the number of the first chosen problem, $b$ is the number of the second chosen problem and $c$ is the number of the third chosen problem.
Examples
input
6 3 1 4 1 5 9
output
4 1 3
input
5 1 1000000000 1 1000000000 1
output
-1 -1 -1
input
9 10 10 11 10 10 10 10 10 1
output
9 8 3
Solution:
private fun readLn() = readLine()!! // string line
private fun readInt() = readLn().toInt() // single int
private fun readLong() = readLn().toLong() // single long
private fun readDouble() = readLn().toDouble() // single double
private fun readStrings() = readLn().split(" ") // list of strings
private fun readInts() = readStrings().map { it.toInt() } // list of ints
private fun readLongs() = readStrings().map { it.toLong() } // list of longs
private fun readDoubles() = readStrings().map { it.toDouble() } // list of doubles
fun main(args: Array<String>) {
var n = readInt()
var a = readInts()
for (i in 0..n-1) {
var j = 0
var k = 0
while (j < n && a[j] >= a[i]) j++
while (k < n && a[k] <= a[i]) k++
if (j < n && k < n) {
println("${j+1} ${i+1} ${k+1}")
return
}
}
println("-1 -1 -1")
}
