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 Ruby (judge verdict: Wrong Answer)

class RMQ
  DEFAULT_VALUE = 2**31 - 1
  def initialize
    @array = []
  end

  def find(x, y)
    @array.each do |h|
      if x <= h[:index] && h[:index] <= y
        puts h[:value]
        return
      end
    end
    puts DEFAULT_VALUE
  end

  def update(x, y)
    @array << { index: x, value: y }
    @array.sort_by! { |h| h[:value] }
  end
end

n, q = gets.split.map(&:to_i)
rmq = RMQ.new
q.times do
  com, x, y = gets.split.map(&:to_i)
  com == 0 ? rmq.update(x, y) : rmq.find(x, y)
end