Problem:
Score : 100 points

Problem Statement

You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types.

The type of i-th query is represented by T_i.

T_i=1: You are given two integers X_i,V_i. Replace the value of A_{X_i} with V_i.

T_i=2: You are given two integers L_i,R_i. Calculate the maximum value among A_{L_i},A_{L_i+1},\cdots,A_{R_i}.

T_i=3: You are given two integers X_i,V_i. Calculate the minimum j such that X_i \leq j \leq N, V_i \leq A_j. If there is no such j, answer j=N+1 instead.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq A_i \leq 10^9

1 \leq Q \leq 2 \times 10^5

1 \leq T_i \leq 3

1 \leq X_i \leq N, 0 \leq V_i \leq 10^9 (T_i=1,3)

1 \leq L_i \leq R_i \leq N (T_i=2)

All values in Input are integer.

Input

Input is given from Standard Input in the following format:

N Q
A_1 A_2 \cdots A_N
First query
Second query
\vdots
Q-th query

Each query is given in the following format:

If T_i=1,3,

T_i X_i V_i

If T_i=2,

T_i L_i R_i

Output

For each query with T_i=2, 3, print the answer.

Sample Input 1

5 5
1 2 3 2 1
2 1 5
3 2 3
1 3 1
2 2 4
3 1 3

Sample Output 1

3
3
2
6

First query: Print 3, which is the maximum of (A_1,A_2,A_3,A_4,A_5)=(1,2,3,2,1).

Second query: Since 3>A_2, j=2 does not satisfy the condition．Since 3 \leq A_3, print j=3.

Third query: Replace the value of A_3 with 1. It becomes A=(1,2,1,2,1).

Fourth query: Print 2, which is the maximum of (A_2,A_3,A_4)=(2,1,2).

Fifth query: Since there is no j that satisfies the condition, print j=N+1=6.

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

class SegmentTree
  def initialize(arg, e)
    if arg.is_a?(Integer)
      v = Array.new(arg){ e }
    elsif arg.is_a?(Array)
      v = arg
    end

    @e  = e
    @op = Proc.new { yield(_1, _2) }

    @n = v.size
    @log = ceil_pow2(@n)
    @leaf_size = 1 << @log
    @d = Array.new(@leaf_size * 2) { e }
    v.each_with_index { |v_i, i| @d[@leaf_size + i] = v_i }
    (@leaf_size - 1).downto(1) { |i| update(i) }
    self
  end

  def set(q, x)
    q += @leaf_size
    @d[q] = x
    1.upto(@log) { |i| update(q >> i) }
  end

  def get(q)
    d[leaf_size + q]
  end

  def prod(l, r)
    sml = @e
    smr = @e
    l += @leaf_size
    r += @leaf_size

    while l < r
      if l[0] == 1
        sml = @op.call(sml, @d[l])
        l += 1
      end
      if r[0] == 1
        r -= 1
        smr = @op.call(@d[r], smr)
      end
      l /= 2
      r /= 2
    end

    @op.call(sml, smr)
  end

  def all_prod
    d[1]
  end

  def max_right(l)
    return @n if l == @n

    f = Proc.new { yield(_1) }

    l += @leaf_size
    sm = @e
    loop do
      break
      l /= 2 while l.even?
      unless f.call(@op.call(sm, @d[l]))
        while l < @leaf_size
          l *= 2
          if f.call(@op.call(sm, @d[l]))
            sm = op(sm, @d[l])
            l += 1
          end
        end
        return l - @leaf_size
      end
      sm = @op.call(sm, @d[l])
      l += 1
      break if (l & -l) == l
    end

    @n
  end

  private

  def update(k)
    @d[k] = @op.call(@d[2 * k], @d[2 * k + 1])
  end
end

def ceil_pow2(n)
  x = 0
  x += 1 while (1 << x) < n
  x
end

SegTree = SegmentTree
Segtree = SegmentTree

INF = (1 << 60) - 1

n, q = gets.to_s.split.map{ |t| t.to_i }
a    = gets.to_s.split.map{ |t| t.to_i }

st = Segtree.new(a, -INF) { [_1, _2].max }

q.times do
  query = gets.to_s.split.map{ |t| t.to_i }
  if query[0] == 1
    _, x, v = query
    st.set(x - 1, v)
  elsif query[0] == 2
    _, l, r = query
    puts st.prod(l - 1, r)
  else
    _, x, v = query
    puts st.max_right(x - 1) { _1 < v } + 1
  end
end