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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

"""
セグメント木
func：二項演算の関数（モノイドである必要あり）
e：単位元（モノイドにおける）
update, find, bisect, 全てにおいて, 1-indexとなっている。
(大抵の場合、Atcoderの問題文の表記のままの値を、メソッドに代入すれば良い)
"""
class SegmentTree:
    def __init__(self, n, func, e, arrange=None):
        self.init(n)
        self.func = func
        self.e = e
        self.make_arrange(arrange)
    
    def init(self, n):
        self.inf = pow(2, 32)
        self.n = n
        self.N = 1
        while self.N < self.n:
            self.N *= 2
        self.size = self.N * 2 - 1
        self.N -= 1
    
    def make_arrange(self, arrange):
        self.set_arrange(arrange)
        self.construct(arrange)

    def set_arrange(self, arrange):
        if arrange == None:
            self.segment = [self.e]*(self.size)
            return
        self.segment = [0]*(self.N) + arrange + [self.e]*(self.size-self.N-self.n)
    
    def construct(self, arrange):
        if arrange == None:
            return
        for i in range(self.N-1, -1, -1):
            self.segment[i] = self.func(self.segment[2*i+1], self.segment[2*i+2])
    
    def update(self, i, x):
        i += (self.N-1)
        self.segment[i] = x
        while i > 0:
            i = (i-1)//2
            self.segment[i] = self.func(self.segment[2*i+1], self.segment[2*i+2])
    

    def count(self, l, r):
        res = self.e
        l += self.N-1
        r += self.N-2
        while r - l:
            if l & 1:
                res = self.func(res, self.segment[l])
                l += 1
            if r & 1:
                r -= 1
                res = self.func(res, self.segment[r])
            l >>= 1
            r >>= 1
        return res
    
    """
    二分探索
    これは区間[l, r]にてx以上となる最小のインデックスを返すメソッド
    まずfind_l：二分木左側を探索し、そこに条件を満たすものがなかったらfind_：右側を探索
    インデックスの最大最小により左右を使い分ける
    """

    def bisect_sub(self, a, b, k, l, r, x):
        if r <= a or b <= l:
            return b+1
        if self.segment[k] < x:
            return b+1
        if k >= self.N:
            return r
        
        find_l = self.bisect_sub(a, b, 2*k+1, l, (l+r)//2, x)
        if find_l <= b:
            return find_l

        find_r = self.bisect_sub(a, b, 2*k+2, (l+r)//2, r, x)
        return find_r

    def bisect(self, l, r, x):
        return self.bisect_sub(l-1, r, 0, 0, self.size-self.N, x)
    

def main():
    n, q = map(int, input().split())
    p = list(map(int, input().split()))
    seg = SegmentTree(n, max, 0, arrange=p)

    res = []
    for i in range(q):
        a, b, c = list(map(int, input().split()))
        if a == 1:
            seg.update(b, c)
        elif a == 2:
            ans = seg.count(b, c)
            res.append(ans)
        else:
            ans = seg.bisect(b, n, c)
            res.append(ans)

    print(*res, sep="\n")
            

    
if __name__ == "__main__":
    main()