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 C++ (GCC 9.2.1 with AC Library v1.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#include <atcoder/all>
using namespace atcoder;
#define rep(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define rep2(i, s, n) for (ll i = s; i < (ll)(n); i++)
#define repr(i, n) for (ll i = n; i >= 0; i--)
#define pb push_back
#define COUT(x) cout << (x) << "\n"
#define COUTF(x) cout << setprecision(15) << (x) << "\n"
#define ENDL cout << "\n"
#define DF(x) x.erase(x.begin())
#define ALL(x) x.begin(), x.end()
#define SORT(x) sort(ALL(x))
#define RSORT(x) sort(x.rbegin(), x.rend())
#define REVERSE(x) reverse(ALL(x))
#define MAX(x) *max_element(ALL(x))
#define MAXI(x) max_element(ALL(x)) - x.begin()
#define SUM(x) accumulate(ALL(x), 0)
#define COUNT(x, y) count(ALL(x), y);
#define ANS cout << ans << "\n"
#define YES cout << "YES\n";
#define NO cout << "NO\n";
#define Yes cout << "Yes\n";
#define No cout << "No\n";
#define YN YES else NO
#define yn Yes else No
#define init() \
  cin.tie(0);  \
  ios::sync_with_stdio(false)
#define LINE cerr << "[debug] line: " << __LINE__ << "\n";
#define debug(x) cerr << "[debug] " << #x << ": " << x << "\n";
#define debugV(v)                       \
  cerr << "[debugV] " << #v << ":";     \
  rep(z, v.size()) cerr << " " << v[z]; \
  cerr << "\n";
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using mll = map<ll, ll>;
using qll = queue<ll>;
using P = pair<ll, ll>;
using vp = vector<P>;
using vs = vector<string>;

template <typename T>
inline istream& operator>>(istream& i, vector<T>& v) {
  rep(j, v.size()) i >> v[j];
  return i;
}

template <typename T1, typename T2>
inline istream& operator>>(istream& i, pair<T1, T2>& v) {
  return i >> v.first >> v.second;
}

constexpr ll INF = 0x3f3f3f3f3f3f3f3f;
constexpr ld PI = 3.141592653589793238462643383279;
ll get_digit(ll x) {
  return to_string(x).size();
}

ll gcd(ll x, ll y) {
  return y ? gcd(y, x % y) : x;
}

ll lcm(ll a, ll b) {
  return a / gcd(a, b) * b;
}

template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}

ll op(ll a, ll b) {
  return max(a, b);
}

ll e() {
  return 0;
}

int target;
bool f(ll x) {
  return x <= target;
}

signed main() {
  init();
  ll N, Q;
  cin >> N >> Q;

  segtree<ll, op, e> seg(N);

  rep(i, N) {
    ll a;
    cin >> a;
    seg.set(i, a);
  }
  rep(i, Q) {
    ll t, l, r;
    cin >> t >> l >> r;
    l--;
    r--;

    if (t == 1) seg.set(l, r);
    if (t == 2) COUT(seg.prod(l, r));
    if (t == 3) {
      target = r;
      COUT(seg.max_right<f>(l) + 1);
    }
  }

  return 0;
}