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: Wrong Answer)

import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


### セグメント木
ninf = -10**9
op = max
class SegmentTree:
    def __init__(self, n, a=None):
        """初期化
        num : n以上の最小の2のべき乗
        """
        num = 1
        while num<=n:
            num *= 2
        self.num = num
        self.seg = [ninf] * (2*self.num-1)
        if a is not None:
            # O(n)で初期化
            assert len(a)==n
            for i in range(n):
                self.seg[num-1+i] = a[i]
            for k in range(num-2, -1, -1):
                self.seg[k] = op(self.seg[2*k+1], self.seg[2*k+2])
    def update(self,i,x):
        """update(i,x):Aiをxに更新する
        """
        k = i+(self.num-1)
        self.seg[k] = x
        k = (k-1)//2
        while k >= 0:
            self.seg[k] = op(self.seg[2*k+1], self.seg[2*k+2])
            k = (k-1)//2
    def query(self,a,b):
        k = 0
        l = 0
        r = self.num
        q = [(k,l,r)]
        ans = ninf
        # 重なる区間を深さ優先探索
        while q:
            k,l,r = q.pop()
            if r<=a or b<=l:
                pass
            elif a<=l and r<=b:
                ans = op(ans, self.seg[k])
            else:
                q.append((2*k+1,l,(l+r)//2))
                q.append((2*k+2,(l+r)//2,r)) 
        return ans
    def find_right(self,a,b,x=None,f=None):
        """[a,b)で値がx以上のインデックスの最大
        存在しない場合-1を返す
        """
        if f is None:
            f = lambda y: y>=x
        k = 0
        l = 0
        r = self.num
        q = [(k,l,r,True)] # 行きがけかどうか
        ans = -1
        while q:
            k,l,r,flg = q.pop()
            if flg:
                if not f(self.seg[k]) or r<=a or b<=l: # 条件を満たせない or 区間が重複しない
                    pass
                elif k>=self.num-1: # 自身が葉
                    ans = max(ans, k - (self.num-1))
                    return ans
                else:
                    # 左への探索を予約
                    q.append((2*k+1,l,(l+r)//2,False))
                    # 右への探索
                    q.append((2*k+2,(l+r)//2,r,True))
            else:
                if ans>=0:
                    return ans
                q.append((k,l,r,True))
        return ans
    def find_left(self,a,b,x=None, f=None):
        """[a,b)で値がx以上のインデックス(0,1,...,self.num-1)の最小
        条件を満たすものが存在しないとき、self.numを返す
        """
        if f is None:
            f = lambda y: y>=x
        k = 0
        l = 0
        r = self.num
        q = [(k,l,r,True)] # 行きがけかどうか
        ans = self.num
        while q:
            k,l,r,flg = q.pop()
            if flg:
                if not f(self.seg[k]) or r<=a or b<=l: # x以上を満たせない or 区間が重複しない
                    continue
                elif k>=self.num-1: # 自身が葉
                    ans = min(ans, k - (self.num-1))
                    return ans
                else:
                    # 右への探索を予約
                    q.append((2*k+2,(l+r)//2,r,False))
                    # 左への探索
                    q.append((2*k+1,l,(l+r)//2,True))
            else:
                if ans<self.num:
                    return ans
                q.append((k,l,r,True))
        return ans
    def query_index(self,a,b,k=0,l=0,r=None):
        """query(a,b,0,0,num):[a,b)の最大値
        最大値を与えるインデックスも返す
        """
        if r is None:
            r = self.num
        if r <= a or b <= l:
            return (ninf, None)
        elif a <= l and r <= b:
            return (self.seg[k], self._index(k))
        else:
            return op(self.query_index(a,b,2*k+1,l,(l+r)//2),self.query_index(a,b,2*k+2,(l+r)//2,r))
    def _index(self, k):
        if k>=self.num:
            return k - (self.num-1)
        else:
            if self.seg[2*k+1]>=self.seg[2*k+2]:
                return self._index(2*k+1)
            else:
                return self._index(2*k+2)

n,q = list(map(int, input().split()))
a = list(map(int, input().split()))
a = a[::-1]
sg = SegmentTree(n, a)
for _ in range(q):
    t,x,y = map(int, input().split())
    if t==1:
        sg.update(n-1-(x-1),y)
    elif t==2:
        print(sg.query(n-1-y, n-1-(x-1)))
    else:
        res = sg.find_right(0,n-1-(x-1)+1,y)
        if res<0:
            print(n+1)
        else:
            print(n-1-res+1)