#include <fstream>
#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
#define inf INT_MIN
using namespace std;
int segm_tree[270000];
int a[100002];

int cont;

void build (int node, int left, int right) { // node - indexul segm_tree, st / dr - capetele segmentului curent
 cont++;
    if (left == right) {
            segm_tree[node] = a[left]; // nod frunza
    } else {
        int middle = (left + right) / 2;
        build (2 * node + 1, left, middle); // apeleaza recursiv ambele jumatati
        build (2 * node + 2, middle +1, right);
        segm_tree[node] = max(segm_tree[2 * node + 1], segm_tree[2 * node + 2]); // recalculare
        }
    }

void update (int node, int left, int right, int x, int y) {
if (left == right)  { // suntem in nodul x (Frunza)
        segm_tree[node] = y;
     } else {
         int middle = (left + right) / 2;
         if (x <= middle) { // actualizam fiul din stanga
            update(2 * node +1, left, middle, x, y);
         } else {
            update(2 * node +2, middle + 1, right, x, y); // actualizam fiul din dreapta
         }
          segm_tree[node] = max(segm_tree[2 * node + 1], segm_tree[2 * node + 2]); // recalculare
     }


}

int query (int node, int left, int right, int x, int y) {
   if (x <= left && right <= y) { // segmentul nodului e complet inclus in query
       return segm_tree[node];
   } else {
       int answer = inf;
       int middle = (left + right) / 2;
       if (x <= middle) { // fiul din stanga si segmentul din query au cel putin un element in comun
          answer = max(answer, query(2 * node + 1, left, middle, x, y));
       }
       if (y >= middle + 1) { // fiul din dreapta si segmentul din query au cel putin un element in comun
           answer = max(answer,query(2 * node + 2, middle + 1, right, x, y));
       }
       return answer;
    }
}

int main()
{
    ifstream f("arbint.in");
    ofstream g("arbint.out");
    int n, m, i, x, y, z;
    f >> n >> m;
    for (i=1; i<=n; ++i)
        f >> a[i];
        build(0,1,n);
        for (i=1; i<=m; ++i)
        {
            f >> x >> y >> z;
            if (x) update(0,1,n,y,z);
            else g << query(0,1,n,y,z) <<"\n";
        }


    return 0;
}