Problem:
Range Minimum Query (RMQ)

Write a program which manipulates a sequence A = {a0, a1, . . . , an-1} with the following operations:

find(s, t): report the minimum element in as, as+1, . . . ,at.

update(i, x): change ai to x.

Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Input

n q
com0 x0 y0
com1 x1 y1
...
comq−1 xq−1 yq−1

In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes update(xi, yi) and '1' denotes find(xi, yi).

Output

For each find operation, print the minimum element.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 100000

If comi is 0, then 0 ≤ xi < n, 0 ≤ yi < 231-1.

If comi is 1, then 0 ≤ xi < n, 0 ≤ yi < n.

Sample Input 1

3 5
0 0 1
0 1 2
0 2 3
1 0 2
1 1 2

Sample Output 1

1
2

Sample Input 2

1 3
1 0 0
0 0 5
1 0 0

Sample Output 2

2147483647
5

Solution in Scala (judge verdict: Accepted)

import java.util.Scanner

object Main extends App { 
  val sc = new Scanner(System.in)
  val n, q = sc.nextInt()
  val list = (for (i <- 0 until q) yield (sc.nextInt(), sc.nextInt(), sc.nextInt())).toList
 
  type Cmd = (Int, Int, Int)

  val df = 2147483647
  val an = Array.fill(n)(df)

  def update(i:Int, x:Int):Unit = an(i) = x
  def find(s:Int, t:Int):Int = {
    def iter(i:Int, m:Int):Int = {
      if (i > t) m 
      else iter(i+1, m min an(i))
    }
    iter(s, df)
  }

  def solver(l:List[Cmd], acc:List[Int]):List[Int] = {
    if (l.isEmpty) acc.reverse
    else {
      val (com, x, y) = l.head
      com match {
        case 0 => { update(x, y); solver(l.tail,acc) }
        case _ => solver(l.tail, find(x, y)::acc)
      }
    }
  }

  solver(list, Nil).foreach(println)
}