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)

class SegmentTree():
    '''
    非再帰
    segment tree
    '''

    def __init__(self, n, func, init=float('inf')):
        '''
        n->配列の長さ
        func:func(a,b)->val,　func=minだとRMQになる
        木の高さhとすると,
        n:h-1までのノード数。h段目のノードにアクセスするために使う。
        data:ノード。
        parent:k->child k*2+1とk*2+2
        '''
        self.n = 2**(n-1).bit_length()
        self.init = init
        self.data = [init]*(2*self.n)
        self.func = func

    def set(self, k, v):
        '''
        あたいの初期化
        '''
        self.data[k+self.n-1] = v

    def build(self):
        '''
        setの後に一斉更新
        '''
        for k in reversed(range(self.n-1)):
            self.data[k] = self.func(self.data[k*2+1], self.data[k*2+2])

    def update(self, k, a):
        '''
        list[k]=aに更新する。
        更新ぶんをrootまで更新
        '''
        k += self.n-1
        self.data[k] = a

        while k > 0:
            k = (k-1)//2
            self.data[k] = self.func(self.data[k*2+1], self.data[k*2+2])

    def query(self, l, r):
        '''
        [l,r)のfuncを求める
        '''
        L = l+self.n
        R = r+self.n
        ret = self.init
        while L < R:
            if R & 1:
                R -= 1
                ret = self.func(ret, self.data[R-1])
            if L & 1:
                ret = self.func(ret, self.data[L-1])
                L += 1
            L >>= 1
            R >>= 1
        return ret

    def lower_bound_index(self,x,v):
        x+=self.n
        while self.L[x]<v:
            if x%2==1:
                if len(bin(x))==len(bin(x+1)):
                    x+=1
                else:
                    return self.n

            else:
                x>>=1
        while x<self.n:
            if self.data[2*x]>=v:
                x=2*x
            else:
                x=2*x+1
        return x-self.n



N,Q = map(int,input().split())
Seg = SegmentTree(N,lambda x,y:max(x,y),init=-1)
A = list(map(int,input().split()))
for i,a in enumerate(A):
    Seg.set(i,a)



Seg.build()

for _ in range(Q):
    t,a,b = map(int,input().split())
    if t==1:
        Seg.update(a-1,b)
    elif t==2:
        print(Seg.query(a-1,b))
    elif t==3:
        print(min(Seg.lower_bound_index(a-1,b),N)*1)