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: Time Limit Exceeded)

MAX = 2**31-1
n, q = gets.split.map(&:to_i)
b = Math.sqrt(n).floor
a = Array.new(n){MAX}
ba = []
n.times{|i|
  j = i / b
  ba[j] = MAX
}

q.times{
  com, x, y = gets.split.map(&:to_i)
  if com == 1 then
    if y - x >= n then
      xb = x / b
      yb = y / b
      min = [MAX, MAX, MAX]
      min[0] = ba[xb+1..yb-1].min if xb + 1 <= yb-1
      min[1] = a[x..xb+b-1].min
      min[2] = a[yb*b..y].min
      p min.min
    else
      p a[x..y].min
    end
  else
    a[x] = y
    j = x / b
    s = j * b
    e = j * b + b
    ba[j] = a[s..e].min
  end
}