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 SegTree
    def initialize(init_ary, lim, &block)
        n = init_ary.size
        @m = 1 << (n-1).bit_length
        @n = @m << 1

        @lim = lim
        @ary = Array.new(@n, @lim)
        @fanc = block

        init_ary.each_with_index {|e,i| @ary[@m+i] = e}
        (@m-1).downto(1) {|i| @ary[i] = @fanc.call(@ary[i*2], @ary[i*2+1])}
    end

    def set(i, x)
        i += @m
        @ary[i] = x
        i >>= 1

        while i >= 1
            @ary[i] = @fanc.call(@ary[i*2], @ary[i*2+1])
            i >>= 1
        end
    end

    def find(i, a, b)
        return @lim if @t < a || b < @s
        return @ary[i] if @s <= a && b <= @t

        m = (a + b) >> 1
        return @fanc.call(find(i*2, a, m), find(i*2+1, m+1, b))
    end

    def find_start(s, t)
        @s = s
        @t = t
        find(1, 0, @m-1)
    end

    def calc3(i, a, b)
        return @lim if @t < a || b < @s || @ary[i][0] < @v
        return @ary[i] if a == b

        m = (a + b) >> 1
        x = calc3(i*2, a, m)
        return x if x[1] != @lim[1]
        return calc3(i*2+1, m+1, b)
    end

    def calc3_start(s, t, v)
        @s = s
        @t = t
        @v = v
        calc3(1, 0, @m-1)
    end
end

n, q = gets.split.map(&:to_i)
a = gets.split.map.with_index{|e,i| [e.to_i, i]}

st = SegTree.new(a, [0, n]) {|ch1, ch2| ch1[0] > ch2[0] ? ch1 : ch2}

q.times do
    t, x, v = gets.split.map(&:to_i)
    if t == 1
        st.set(x-1, v)
    elsif t == 2
        puts st.find_start(x-1, v-1)[0]
    else
        puts st.calc3_start(x-1, n-1, v)[1] + 1
    end
end