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)

n, q = gets.split.map(&:to_i)
b = Math.sqrt(n).floor

a = Array.new(n){2**31-1}
ba = Array.new(n/b){2**31-1}

q.times{
  com, x, y = gets.split.map(&:to_i)
  case com
  when 0
    a[x] = y
    ba[x/b] = y if ba[x/b] > y
  when 1
    min = 2**31-1
    bx = x / b
    by = y / b
    min = 2**31-1
    x.upto(bx*b+b-1){|j|
      min = a[j] if min > a[j]
    }
    (bx+1).upto(by-1){|j|
      min = ba[j] if min > ba[j]
    }
    (by*b).upto(y){|j|
      min = a[j] if min > a[j]
    }
    p min
  end
}