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: Accepted)

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

    def initialize(v, e, &op)
        @n = v.size
        @log = (@n - 1).bit_length
        @size = 1 << @log
        @op = op

        @d = Array.new(2 * @size, e) # vector

        @n.times{|i| @d[@size + i] = v[i]}
        (@size - 1).downto(1) {|i| update(i)}
    end

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

    def get(pos); @d[pos + @size]; end

    def prod(l, r)
        sml = smr = @d[0]
        l += @size
        r += @size

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

    def all_prod; @d[1]; end

    def max_right(l, &f)
        return @n if l == @n
        l += @size
        sm = @d[0]

        loop do
            l >>= 1 while l[0] == 0
            x = @op.call(sm, @d[l])
            if !f.call(x)
                while l < @size
                    l <<= 1
                    x = @op.call(sm, @d[l])
                    if f.call(x)
                        sm = x
                        l += 1
                    end
                end
                return l - @size
            end
            
            sm = x
            l += 1
            break if (l & -l) == l
        end

        @n
    end
    
    def min_left(r, &f)
        return 0 if r == 0
        r += @size
        sm = @d[0]

        loop do
            r -= 1
            r >>= 1 while r > 1 && r[0] == 1
            x = @op.call(@d[r], sm)
            if !f.call(x)
                while r < @size
                    r = 2 * r + 1
                    x = @op.call(@d[r], sm)
                    if f.call(x)
                        sm = x
                        r -= 1
                    end
                end
                return r + 1 - @size
            end
            sm = x
            break if (r & -r) == r
        end

        0
    end
end

n, q = gets.split.map(&:to_i)
st = SegTree.new(gets.split.map(&:to_i), -1) {|ch1, ch2| ch1 > ch2 ? ch1 : ch2}

q.times do
    t, x, y = gets.split.map(&:to_i)
    
    if t == 1
        st.set(x-1, y)
    elsif t == 2
        puts st.prod(x-1, y)
    else
        puts st.max_right(x-1){|v| v < y} + 1
    end
end