#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define int long long
#define oset tree < pair<int, int> ,  null_type ,  less<pair<int, int>> ,  rb_tree_tag ,  tree_order_statistics_node_update >
using namespace std;
set<int> subarrays;
int findans(int x){ //The second type of query
    //For any element k, the first element in the subarray of element k is -
    //The largest element j in sa such j<=k

    //Find the first element j such that j > k
    //j--; (The previous element)
    auto it = subarrays.upper_bound(x);
    it--;
    return (*it);
}
void remove(int x){
    subarrays.erase(x);
}
signed main(){
    //N -> size of the array// N<=1e6
    //A[N] positive integers
    //We are dividing the entire array into subarrays
    //If i and i+1 are in the same subarray then A(i+1)%A(i) == 0
    //S(i) is the index of the subarray in which the ith element lies
    //Q queries, Q = 2e5
    //Changing A(i) to X
    //2 i, we have to output the first element in the same subarray as i
    //Required complexity is either O(Q) or O(Q.LogN)

    //In this problem, we have to form continuous subarrays
    //For any index i, if A(i+1)%A(i) ==0, then we should place i+1 and i in the same subarray
    //If we can place A(i), A(i+1) in the same subarray, then they should be in the same subarray

    //How do we store the partitions of the subarrays?

    //Since the subarrays are all contigious, we only need to store the first element of each subarray

    //{2, 6, 5, 10, 20, 3}
    //{2, 6} -> index of the first element of each subarray
    //{5, 10}
    //{3}

    //sa
    //{1 -> 1/2, 3 -> 3/4/5, 6 -> 6}
    //For the second type of query, we need to output the first element of the subarray of index k
    //For any element k, the first element in the subarray of element k is -
    //The largest element j in sa such j<=k

    //We need a data structures that can:
    //Store elements in a sorted order - set
    //Erase elements in O(1) or O(LogN) - set
    //Find the largest element j in sa such j<=k in O(1) or O(LogN) - set

    int n, q;
    cin>>n>>q;
    int a[n+1];
    for(int i = 1;i<=n;i++) cin>>a[i];
    //We need to create the subarrays
    subarrays.insert(1); //1 will always be the start of some subarray
    for(int i = 2;i<=n;i++){
        //For any index i, if A(i+1)%A(i) ==0, then we should place i+1 and i in the same subarray
        //If we can place A(i), A(i+1) in the same subarray, then they should be in the same subarray
        if(a[i]%a[i-1] == 0) continue; //we can place i and i-1 in the same subarray
        //we skip because we have already stored the start of the (i-1)th subarray
        subarrays.insert(i); //When a[i] and a[i-1] could not be in the same subarray, we inserted i into the set
    } //We have created the basic partitions before any queries took place

    while(q--){
        int type, i;
        cin>>type>>i;
        if(type==1){
            cin>>a[i]; //the updated value of a(i)
            //There are 4 things that can happen
            //a(i) and a(i-1) can now be in the same subarray -> DONE
            //A(i) and A(i-1) can no longer be in the same subarray -> DONE
            //A(i) and A(i+1) can now be in the same subarray -> DONE
            //A(i) and A(i+1) can no longer be in the same subarray
            if(i>1) {
                if ((findans(i) == i) &&
                    ((a[i] % a[i - 1]) == 0)) { //now a[i] and a[i-1] should be in the same subarray
                    //We need to reverse the operation on line 68
                    remove(i);
                } else if ((findans(i) < i) &&
                           ((a[i] % a[i - 1]) != 0)) { //this means a[i] and a[i-1] were in the same subarray
                    //We need to perform the operation on line 68 to separate the elements
                    subarrays.insert(i);
                }
            }
            if(i<n){
                if((findans(i+1)>i && ((a[i+1] % a[i]) == 0))){
                    remove(i+1);
                }
                else if((findans(i+1)<=i) && ((a[i+1]%a[i])!=0)){
                    subarrays.insert(i+1);
                }
            }
        }
        else{
            cout<<findans(i)<<endl;
        }
    }
}
//acdabcd