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)

class RMQ
    INT_MAX = 2**31-1
    def initialize(n)
    @n = 2**Math::log2(n).ceil   # nが2のべき乗になるように調整
    @dat = Array.new(2*@n-1) { INT_MAX }
    end
    def update(i, x)
        i += @n-1
    @dat[i] = x
    while i > 0                # 葉から親へ辿る処理
          i = (i-1)/2
    @dat[i] = [@dat[i*2+1], @dat[i*2+2]].min
    end
    end
    def find(s, t)
        query(s, t+1, 0, 0, @n)
    end
    private
    def query(a, b, k, l, r)
    return INT_MAX if r <= a || b <= l # 交差しない?
    return @dat[k] if a <= l && r <= b         # 完全に含む?
        vl = query(a, b, k*2+1, l, (l+r)/2)
        vr = query(a, b, k*2+2,(l+r)/2, r)
        [vl, vr].min
    end
    end
    n, q = gets.split.map(&:to_i)
    rmq = RMQ.new(n)
    q.times {
      com, x, y = gets.split.map(&:to_i)
    case com
    when 0
        rmq.update(x, y)
    when 1
    puts rmq.find(x, y)
    end
    }