HackerEarth Operations on an array problem solution

In this HackerEarth Operations on an array problem solution, You are given an array of n elements and an integer x. You must perform the following types of operations on the array:

Find the index of the kth occurrence of x in the range l to r (both inclusive). If there are no indexes that satisfy the condition, then print -1.

Update the value that is present at the given index.

For each query of type 1, print the index of the kth occurrence of x.

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

const int N = 1000000 + 1;
int arr[N];
struct BIT{//use one based indexing
    int N;
    vector<int> bit;

    void init(int n){
        bit.clear();
        N = n + 9;
        bit.assign(n + 10, 0);
    }

    void update(int idx, int val){
        while(idx <= N){
            bit[idx] += val;
            idx += idx & -idx;
        }
    }

    int pref(int idx){
        int ans = 0;
        while(idx > 0){
            ans += bit[idx];
            idx -= idx & -idx;
        }
        return ans;
    }

    int rsum(int l, int r){
        return pref(r) - pref(l - 1);
    }

    int kthOrder(int k){
        if(pref(N - 1) < k){
            return -1;
        }
        int cur = 0,sum = 0, ExtraSize = N;
        int ln = log2(ExtraSize);
        for(int i = ln;i >= 0 ; --i){
            int temp = cur + (1 << i);
            if((temp < ExtraSize) && (sum + bit[temp]) < k){
                cur = temp;
                sum += bit[temp];
            }
        }
        return cur + 1;
    }

    //one based indexing
    int lower_bound(int val){
        return (pref(val - 1) + 1);
    }
    //one based indexing
    int upper_bound(int val){
        return (pref(val) + 1);
    }
}bt;
signed main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    int n,x;
    cin >> n >> x;
    bt.init(n);
    for(int i = 1;i <= n;i++){
        cin >> arr[i];
        if(arr[i] == x){
            bt.update(i,1);
        }
    }
    int q;
    cin >> q;
    while(q--){
        int type;
        cin >> type;
        if(type == 1){
            int l,r,k;
            cin >> l >> r >> k;
            k += bt.pref(l - 1);
            if(k > bt.pref(r)){
                cout << "-1n";;
                continue;
            }
            cout << bt.kthOrder(k) << "n";
        }else{
            int l,val;
            cin >> l >> val;
            if(arr[l] == x){
                bt.update(l,-1);
            }
            if(val == x){
                bt.update(l,1);
            }
            arr[l] = val;
        }
    }
}