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)

MAX = 2**31-1

def rmq(a, b, k, l, r, node)
  return MAX if r <= a || b <= l
  return node[k] if a <= l && r <= b
  m = (l+r)/2
  vl = rmq(a, b, 2*k+1, l, m, node)
  vr = rmq(a, b, 2*k+2, m, r, node)
  return vl < vr ? vl : vr
end

n, q = gets.split.map(&:to_i)

m = 1
while m < n
  m *= 2
end
node = Array.new(2*m-1){MAX}

q.times{
  com, x, y = gets.split.map(&:to_i)
  case com
  when 0
    k = x + m - 1
    node[k] = y
    while k > 0
      k = (k-1)/2
      node[k] = node[2*k+1] < node[2*k+2] ? node[2*k+1] : node[2*k+2]
    end
  when 1
    p rmq(x, y+1, 0, 0, m, node)
  end
}