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: Accepted)

def get_i() #空白区切の入力を数値(整数)の配列で返す
  return gets.chomp.split(" ").map(&:to_i)
end
def get_f() #空白区切の入力を数値(実数)の配列で返す
  return gets.chomp.split(" ").map(&:to_f)
end
def get() #空白区切の入力を文字列の配列で返す
  return gets.chomp.split(" ")
end
def get_nsp() #入力されたものを一文字ずつに区切った文字列の配列で返す
  return gets.chomp.split("")
end
def yn_judge(bool,y="Yes",n="No") #boolに真偽を投げることで、trueならy、falseならnの値を出力する
  return bool ? y : n 
end
def array(size1,init=nil,size2=1) #size2に二次元配列時の最初の要素数、size1に次の配列の大きさ、initに初期値を投げることでその配列を返す
  if size2==1
    return Array.new(size1){init}
  else
    return Array.new(size2){Array.new(size1){init}}
  end
end

def min(a,b)
    return a>b ? b : a
end
class SegTree
    attr_accessor:node,:n,:monoid
    def initialize(n,monoid)
        @n=1
        @n*=2 while @n<n
        @monoid=monoid
        @node=array(2*@n,@monoid)
    end

    def update(i,x)
        i+=@n
        @node[i]=x
        while i>0
            i/=2
            @node[i]=min(@node[i*2],@node[i*2+1])
        end
    end

    def get_min(l,r)
      l+=@n
      r+=@n
      res=@monoid
      while l<r
        if l.odd?
            res=min(res,@node[l])
            l+=1
        end 
        if r.odd?
            r-=1
            res=min(res,@node[r])
        end
        l/=2
        r/=2
      end
      return res
    end
end

N,Q=get_i
st=SegTree.new(N,2**31-1)
Q.times do
    c,x,y=get_i
    if c.zero?
      st.update(x,y)
    else
      puts st.get_min(x,y+1) 
    end
end